Instruction:
If  a thiazide diuretic is administered for hypertension, the response seen within 24 hrs on the excretion of different electrolytes are:
A. Sodium(Upward arrow) Potassium (Upward arrow) Calcium(Upward arrow)
B. Sodium(Upward arrow) Potassium(Downward arrow) Calcium(Upward arrow)
C. Sodium(Upward arrow) Potassium(Downward arrow) Calcium(Upward arrow)
D. Sodium(Upward arrow) Potassium(Upward arrow) Calcium(Downward arrow)

Response:
These are medium efficacy diuretics with a primary site of action in the cortical diluting segment or the early DT (Site III). Here they inhibit Na+-Cl symport at the luminal membrane. They do not affect the corticomedullary osmotic gradient indicating lack of action at the medullary thick AscLH.
Under their action, an increased amount of Na is presented to the distal nephron, more of it is exchanges with K+, urinary K+ excretion increased in parallel to the natriuretic response. They decrease renal Ca excretion and increase Mg excretion by a direct distal tubular action. They also decrease urate excretion by the same mechanism as furosemide.

So the answer to this question is D


Instruction:
Struma ovarii is a
A. Metastatic tumor
B. Teratoma
C. Sex cord tumor
D. Surface epithelial tumor

Response:
Struma is a river in europe,along its banks endemic goitre was noted. Strauma is loosely used for goitre in europe. A Struma ovarii benign ovarian tumor is a form of teratoma that contains mostly thyroid tissue, which may develop thyrotoxicosis. Source : Bailey and love 26th edition Pg : 748

So the answer to this question is B


Instruction:
All are features of fetal distress except :
A. Meconium staining of the liquor amnii
B. Hea rate less than 100
C. Involuntary muscle movements
D. Fetal scalp blood pH 7.3

Response:
Fetal scalp blood pH 7.3

So the answer to this question is D


Instruction:
A 19-year-old presents at the emergency department with high fever, severe headache, nausea, and stiff neck for 3 days. The attending physician suspects meningitis and obtains a sample of CSF using a lumbar puncture. From which of the following spaces was the CSF collected?
A. Epidural space
B. Subdural space
C. Subarachnoid space
D. Pretracheal space

Response:
The subarachnoid space, containing the CSF, is located between the pia and the arachnoid mater. Neither the epidural space, the subdural space, nor the pretracheal space contains CSF. Although the central canal, contained within the substance of the spinal cord, does contain CSF, extraction of CSF from this region would result in spinal cord injury. CSF circulates in the area of the subarachnoid space and can be aspirated only from that location. The subdural space is only a potential space between the dura and arachnoid mater. The epidural space contains the epidural fat and Batson's venous plexus and is the preferred site for aspirating CSF for diagnostic purposes (and epidural anesthesia). CSF is not located in the pretracheal space.

So the answer to this question is C


Instruction:
An 8 year old child presented with signs of increased intracranial tension. Imaging studies showed a growth in the brain. Which is the commonest tumor in a child in the posterior fossa of head?
A. Astrocytoma
B. Medulloblastoma
C. Craniopharyngioma
D. Meningioma

Response:
Medulloblastoma arises in the posterior pa of the cerebellar vermis and neuroepithelial roof of the fouh ventricle in children. It accounts for 20 percent of childhood brain tumors. Primitive neuroectodermal tumor is the most common type of medulloblastoma. Most occur in the first decade of life, but there is a second peak around age 30. Medulloblastoma is the most common malignant pediatric brain tumor. They are usually midline. Most occur in the cerebellum and present with symptoms of increased ICP. Ref: Ropper A.H., Samuels M.A. (2009). Chapter 31. Intracranial Neoplasms and Paraneoplastic Disorders. In A.H. Ropper, M.A. Samuels (Eds), Adams and Victor's Principles of Neurology, 9e.

So the answer to this question is B


Instruction:
A child from a small village from bihar has long bone pain, is weak and lethargic and on examination he has bow legs. The X-Ray reports shows increase in bone density, osteophytes and dismorphic joint space. Diagnosis:
A. Fluorosis
B. Rickets
C. Scurvy
D. Caffeys disease

Response:
Ans. (a) FluorosisRef: Maheshwari 5th ed. / 314* Hint in the question is child from a village of Bihar with Genu varum/bow legged, long bone pain and X-Rays showing increased bone density, this characteristically focuses towards Fluorosis.* If its genu valgum (knock knee) then answer is Rickets,GENU VALGUM (KNOCK KNEE)* It is abnormal approximation of knees with abnormally divergent ankles. It can be estimated by measuring the distance between the medial malleolis, when the knees are touching with the patella facing forwards; it is usually <5cm. Valgus alignment of lower extremities is normal in child between 2-8 years of age (Known as physiological valgus & is maximum between 2-4 years). The commonest cause of genu valgum (knock knee) is idiopathic > rickets.GENU VARUM (BOW LEGS)* Knee are abnormally divergent & ankles approximated. B/L bow legs can be estimated by measuring the distance between the medial malleoli when heels are touching; it should be <6cm. Normal children show maximum varus at 6 months to 1 year of age. The causes of genu varum are similar to genu val gum except that the defective growth is on the medial side.FLUOROSIS* Fluorine in very low concentration- 1 part per million (ppm) or less is used to reduce the incidence of dental caries. At slightly higher levels (2-4 ppm) it may produce mottling of teeth.* In some parts of India & Africa, where fluorine concentration in the drinking water may be above 10 ppm- chronic fluorine intoxication (flurosis) is endemic and results in skeletal anomalies.* Characterstic pathological feature is subperiosteal new bone accretion and osteosclerosis (increased bone density), most marked in vertebrae, ribs, pelvis, forearm and leg bones, together with hyperostosis at the bony attachments of ligaments, tendons and fascia.* Despite the apparent thickening & density of skeleton, tensile strength is reduced and the bones fracture more easily.* First clinical manifestation is usually a stress fracture, back pain, bone pain, joint stiffness and neurological defects ( due to hyperostosis encroaching vertebral canal).* Characterstic x- ray features are osteosclerosis, osteophytosis and ossification of ligamentous and fascial attachments. It can be radiologically mistaken for other osteosclerotic conditions as Paget s disease, osteopetrosis, renal osteodystrophy, idiopathic skeletal hyperostosis etc.* NOTE: Fluorine stimulates osteoblastic activity; fluoroapatite crystals are laid down in bone and these are usually resistant to osteoclastic resorption. This leads to calcium retention impaired mineralization and secondary hyperparathyroidism.Also Know* Infantile cortical hyperostosis (Caffeys Disease) is a self limiting disorder characterized by soft tissue swelling, rapid subperiosteal new bone formation, cortical thickening of under lying bones, fever and irritability. Classically, the onset of disease occurs before 5th month of life with resolution by 3 years of age.* In sporadic cases, mandible is the most common site of involvement presenting as jaw tumor or swelling i.e. is firm, tender without heat or redness.

So the answer to this question is A


Instruction:
Which of the following is not the period of physiological raising of the bite according to Martin and Schwarz
A. Eruption of 6th year molar
B. Eruption of 12th year molars
C. Eruption of 18th year molars
D. None of the above

Response:
Bruxism synonym.
Neuralgial traumatic (Karolyl)
Karolyi effect (Weskit)
Occlusal habit neurosis (Tischer)
Parafunctional occlusal habit (Drum)
According to Ramjford and Ash, BRUXISM is triggered by: Emotional tensions Occlusal maladjustment Pain / discomfort

So the answer to this question is D


Instruction:
Which of the following is the most penetration beam –
A. Electron beam
B. 5 MV photons
C. 18MV photons
D. Proton beam

Response:
Gamma rays (high energy photons) have the highest penetrating power. Therefore, 18 MeV photons will have high penetrating power.

So the answer to this question is C


Instruction:
Dofetilide is which class of anti arrhythmia drug
A. Class 1
B. Class 11
C. Class 111
D. Class 1V

Response:
Refer kDT 7/e 529 Dofetilide is a class III antiarrhythmic agent Dofetilide is used for the maintenance of sinus rhythm in individuals prone to the occurrence of atrial fibrillation and flutterarrhythmias, and for chemical cardioversionto sinus rhythm from atrial fibrillation and flutter

So the answer to this question is C


Instruction:
The main function of Vitamin C in the body is -
A. Coenzyme for energy metabolism
B. Regulation of lipid synthesis
C. Involvement as antioxidant
D. Inhibition of cell growth

Response:
Ans. is 'c' i.e., Involvement as antioxidant Vitamin C ( Ascorbic acid)o Ascorbic acid (Vitamin C) is also called antiscorbutic factor. It is very heat labile, especially in basic medium. Ascorbic acid itself is an active form. Maximum amount of vitamin C is found in adrenal cortex,o Ascorbic acid functions as a reducing agent and scavanger of free radicals (antioxidant). Its major functions are:- In collagen synthesis: - Vitamin C is required for post-translational modification by hydroxylation of proline and lysine residues converting them into hydroxyproiine and hydroxy lysine. Thus vitamin C is essential for the conversion of procollagen to collagen, which is rich in hydroxyproiine and hydroxylysine. Through collagen synthesis, it plays a role in formation of matrix of bone, cartilage, dentine and connective tissue.Synthesis of norepineph rine from dopamine by dopamine-b-monoxygenase (dopamineb-hydroxylase) requires Vitamin C.Carnitine synthesisBile acid synthesis :- 7-a-hydroxylase requires vitamin C.Absorption of iron is stimulated by ascorbic acid by conversion of ferric to ferrous ions.During adrenal steroid synthesis, ascorbic acid is required during hydroxylation reactions.Tyrosine metabolism : - Oxidation of P-hydroxy-phenylpyruvate to homogentisate.Folate metabolism : - Folic acid is converted to its active form tetrahydrofolate by help of Vitamin C.

So the answer to this question is C


Instruction:
Which is raised in polycystic ovarian syndrome:
A. 17-alpha-OH-progesterone
B. FSH
C. LH
D. TSH

Response:
Ans. C. Luteinizing hormonePolycystic ovarian syndrome (PCOS) OR disease (PCOD)/stein Leventhal syndromea. PCOS includes chronic non-ovulation and Hyperandrogenemia associated with normal or raised estrogen (E2), raised LH, and low FSH/LH ratio.b. Age -15-25 yearsFeaturesa. The raised E2 level causes negative feedback to pituitary resulting in diminished FSH but raised LH.b. The involvement of adrenal glands is seen in raised androstenedione, dehydroepiandrosterone, testosterone and 17-alpha-Hydroxy-progesterone.c. Much of the testosterone is secreted by the ovarian stroma.d. Macroscopicallye. OvariesBilateral Enlarged LobulatedFree of Adhesion"Necklace" appearance on ultrasoundClinical Featuresa. The women is young who complains of oligomenorrhea and often amenorrhea.b. Infertility occurs in 30%.c. Obesity and hirsutism.DiagnosisLaparoscopic evaluation is not only diagnostic, but also therapeutic.Treatment of PCOSa. Oestrogen suppresses androgen and adrenal production. It is best given with progestogen with no androgenic properties, cyclically as in oral contraceptives.b. Dexamethasone 0.5mg or prednisone 5mg at bedtime also reduces androgen production.c. Hirsutism is treated with cyproterone acetate or spironolactone.d. Infertility is treated with clomiphene.e. In clomiphene failed group, ovulation can be induced with FSH or GnRH analogues.Surgery is reserved for those in whom:a. Medical therapy failsb. Hyperstimulation occursc. Use of GnRH analogues is a cost constraint.d. Surgery comprises laparoscopic multiple punctures of the cysts with electrocautery or laser.

So the answer to this question is C


Instruction:
All the statements are true for medullary carcinoma breast except?
A. Associated with BRCA-2.
B. Usually triple negative
C. Axillary lymph node enlargement bacause of reactive hyperplasia
D. Lymphoreticular infiltration

Response:
Medullary carcinoma of breast Invasive breast cancer 5% of breast cancers Associated with BRCA-1 Axillary lymph node enlargement bacause of reactive hyperplasia Lymphoreticular infiltration Electron M/E - Bizarre cells with high-grade nuclear features Syncytial pattern -Sheets of cells surrounded mononuclear lymphocytes. Borders push into the adjoining breast tissue In its pure form, High grade, Triple negative - ER and PR negative,Her-2-neu negative

So the answer to this question is A


Instruction:
Pott's puffy tumor:
A. Subperiosteal abscess of frontal bone
B. Subperiosteal abscess of ethmoid bone
C. Mucocele of frontal bone
D. Mucocele of ethmoid bone

Response:
Answer. A. Subperiosteal abscess of frontal bonePott's puffy tumor, first described by Sir Percivall Pott in 1760, is a rare clinical entity characterized by subperiosteal abscess associated with osteomyelitis. It is characterized by an osteomyelitis of the frontal bone, either direct or through haematogenous spread. This results in a swelling on the forehead, hence the name. The infection can also spread inwards, leading to an intracranial abscess. Pott's puffy tumor can be associated with coical vein thrombosis, epidural abscess, subdural empyema, and brain abscess.

So the answer to this question is A


Instruction:
Type of collagen forming basement membrane of kidney:
A. I
B. II
C. III
D. IV

Response:
D i.e. IV

So the answer to this question is D


Instruction:
Which among the following will complicate as hyperchloremic acidosis -
A. Ureterosigmoidostomy
B. Diarrhoea
C. Vomiting
D. Ileoplasty

Response:
Ans is 'a' ie Ureterosigmoidostomy

So the answer to this question is A


Instruction:
Most common type of supra condylar fracture in children ?
A. Posteromedial extension
B. Posterolateral extension
C. Anteromedial flexion
D. Anterolateral flexion

Response:
Ans. is 'a' i.e., Posteromedial extension Supracondylar fracture of humerus Supracondylar humeral fractures are the most common elbow fractures in children. Most common age group affected is 5-8 years. Boys are affected more than girls. Left side is more common than right. Mechanism of injury Mostly it occurs due to hyperextension injury. Fracture is caused by a fall onto the outstretched hand with hyperextension at elbow. Types of supracondylar fracture Supracondylar fracture is broadly classified into extension type and flexion type. 1) Extension type It is the most common type (97 - 99%). Distal fragment is extended (tilted backward/posteriorly) in relation to proximal fragment. Occurs due to hyperextension injury after fall on outstretched hands. Generally, displacement of distal fragment may be : - i) Posteromedial (70-80%) ii) Posterolateral (20-30%) 2) Flexion type It is less common type (1-3 %) Distal fragment is flexed (tilted forward/anteriorly) in relation to proximal fragments. The mechanism of injury generally is believed to be a fall directly onto the elbow rather than a fall on outstretched hand. As the extension type fracture is more common (97 - 99%), the most common elbow injury in children is extension type of supracondylar fracture. Clinical features of supracondylar fracture Following a fall, the child is in pain and elbow is swollen. In extension type of injury, `S' shaped deformity of the elbow is obvious. There is loss of both active and passive movements of elbow. Symptoms relating to vascular and nerve injury may be seen. Unusual posterior prominence of the point of elbow (tip of olecrenon) because of backward tilt of the distal fragment. Three point bony relationship is maintained as the fracture is above the level of condyles. Dimple sign due to one of the spikes of proximal fragment penetrating the muscle and tethring the skin.

So the answer to this question is A


Instruction:
A 35 yr old man from Shimoga, Karnataka presents with sudden onset of fever, headache, and myalgia. He also has conjunctivitis and skin bleeds. He recently went to a forest in Shimoga that has a huge population of monkeys. Which of these is the vector for the suspected disease?
A. Ticks
B. Mosquito
C. Fleas
D. Mite

Response:
From the history and symptoms, the most likely diagnosis is Kyasanur forest disease, which is transmitted by ticks. It is a hemorrhagic fever found in Karnataka. These ticks may also act as reservoir. Monkeys act as amplifier hosts. Reference: Ananthanarayan and Paniker&;s Textbook of Microbiology Tenth edition

So the answer to this question is A


Instruction:
The mechanism of action of APF is via following reaction products:
A. Fluoridated hydroxyapatite
B. Stannous trifluorophosphate
C. Dicalcium phosphate dihydrate (DCPD)
D. DCPD + Calcium fluoride + fluorapatite

Response:
When APF is applied on the teeth, it initially leads to dehydration and shrinkage in the volume of hydroxypatite crystals which further on hydrolysis forms an intermediate product called dicalcium phosphate dihydrate (DCPD). This DCPD is highly reactive with fluoride ion and starts forming immediately when APF is applied.
Fluoride penetrates into the crystals more deeply through the openings produced by shrinkage and leads to formation of fluorapatite.
For the conversion of whole of DCPD so formed into fluorapatite, deeper penetration and continuous supply of fluoride is required. Because of this reason, APF is applied every  30 seconds and the teeth has to be kept wet for 4 minutes.
Because high fluoride concentration and low pH favor fluoride deposition, acidification of the fluoride solution with phosphoric acid was found to suppress the dissolution of enamel, as well as the formation of calcium fluoride and provide a more effective treatment.
The intermediate product formed is the dicalcium phosphate dihydrate and calcium fluoride is the principal reaction product.

Essentials of preventive and community dentistry
Soben Peter
5th edition

So the answer to this question is D


Instruction:
In Shigella dysentery associated hemolytic uremic syndrome, the false statement is –
A. Leucocytosis
B. Neurological abnormalities
C. Hepatic failure
D. Thrombotic angiopathy

Response:
Hepatic failure does not occur in HUS.

Laboratory findings of HUS

Thrombocytopenia
ARF →↑BUN and creatinine with oliguria
Peripheral smear Helmet cells, burr cells, fragmental RBCs (Schistocytes).
Hemoglobinuria with hemosiderinuria
Leukocytosis
Microscopic hematuria
Proteinuria
Normal PT and APTT
Option B requires some explanation here -
Though neurological manifestations may occur in HUS, these are not common and this feature differentiated HUS from TTP. In TTP neurological symptom is one of the clinical criteria for diagnosis.

So the answer to this question is C


Instruction:
Most dependent Bronchopulmonary segment in supine position is
A. Left Superior segment
B. Right Superior segment
C. Right posterior basal segment
D. Left posterior basal segment

Response:
Most dependent Bronchopulmonary segment in supine position is Right Superior segment.

So the answer to this question is B


Instruction:
In unconjugated hyperbilirubinemia, the risk of kernicterus increases with the use of –
A. Ceftriaxone
B. Phenobarbitone
C. Ampicillin
D. Sulphonamide

Response:
The administration of sidfonamides to newborn infants, especially if premature, may lead to the displacement of Bilirubin from plasma albumin. In newborn infants, free Bilirubin can become deposited in the basal
subthalamic nuclei of the brain 'causing an encephalopathy called kernicterus. Sulfonamides should not be given to pregnant women near term because these drugs pass through the placenta and are secreted in milk.
Kernicterus is due to unconjugated hyperbilirubinemia.

So the answer to this question is D


Instruction:
True about adrenal pheochromocytoma is
A. Chromaffin negative
B. Mostly malignant
C. Bilateral in 10 percnt cases
D. Unilateral in 10 percnt cases

Response:
Refernce Robbins page no 1134 The "10 percent rule" forpheochromocytoma/paraganglioma--10% familial, 10% malignant, 10% extra-adrenal--is therefore no longer tenable. Current roles of pathology are limited to diagnosing primary or metastatic tumors and identifying features suggestive of malignant potential or hereditary disease.

So the answer to this question is C


Instruction:
Surgical removal of torus is indicated when
A. Torus impinges on soft palate
B. It interferes with development of proper PPS
C. Torus is an undercut that traps debris and food
D. All of the above

Response:
In addition to above indications, torus should also be removed when it interferes with development of adequately extended denture or when patient is too much concerned about it.

So the answer to this question is D


Instruction:
Metronidazole is effective in all of the following conditions except :
A. Pseudomembranous colitis
B. Neurocysticercosis
C. Giardiasis
D. Amebic liver abscess

Response:
None

So the answer to this question is B


Instruction:
The safest drug in pregnancy :
A. INH
B. Rifampicin
C. Ethambutol
D. Streptomycin

Response:
INH

So the answer to this question is A


Instruction:
Onion skin lesion in vessels is seen in ?
A. Malignant hypeension
B. Benign hypeension
C. Peripheral vascular disease
D. None

Response:
Ans. is 'a' i.e., Malignant hypeensionIn malignant hypeension, there is concentric laminated thickening of aeriolar wall, onion skinning.

So the answer to this question is A


Instruction:
Which of the following antenatal complications may cause placentomegaly?
A. Hypeension
B. Diabetes
C. Abruption
D. HIV

Response:
Placentomegaly- Causes Maternal causes: Diabetes Anemia and malnutrition Infections: TORCH , CMV, Parvo virus, syphilis. Fetal causes Hydrops fetalis: Rh incompatibility, ABO incompatibility, fetal cardiac failure Chronic intrauterine infections Umbilical vein obstruction H mole TTTS Placental causes Placental tumours: chorangioma

So the answer to this question is B


Instruction:
Which of the following trace element has vitamin E like action:
A. Selenium
B. Cheomycin
C. Copper
D. Zinc

Response:
Selenium

So the answer to this question is A


Instruction:
Nerve supply of stapedius is:
A. 2nd nerve
B. 3rd nerve
C. 5th nerve
D. 7th nerve

Response:
Ans. is 'd' i.e. 7th C.N. Muscles of the ossicles in middle ear : Stapedius - N.S. is VII CN Tensory Tympanis - N.S. is Mandibular nerve (a br. of trigeminal (V) CN).

So the answer to this question is D


Instruction:
Sparrow marks are seen in?
A. Gunshot injuries
B. Stab injury of face
C. Vitriolage
D. Windshield glass injury

Response:
D i.e. Windshield glass injury Bizarre shaped multiple lacerations on face from contact of shatterd windscreen glassQ of vehicle are known as sparrow foot (patterned) marks.

So the answer to this question is D


Instruction:
Which of the following do not manifest on first day of life?
A. Jaundice
B. Sepsis
C. Neonatal tetanus
D. Meconium aspiration syndrome

Response:
Neonatal tetanus manifests on 8th day of life; thus referred as '8th day disease.' Remaining all conditions given, can manifest on 1st day of life. Also remember, - Neonatal tetanus Elimination criteria : Rate < 0.1 case / 1000 LB. Coverage TT > 90%. Attended deliveries > 75%.

So the answer to this question is C


Instruction:
Type4 Ehlers-Danlos syndrome is characterized by:
A. Deficiencies in hydroxylase and lysyl oxidase.
B. Decreased amount of type III collagen.
C. Nonfunctioning plasma fibronectin.
D. All of the above.

Response:
In EDS types I and II, the classic variety, causative mutations may involve the COL5A1, COL5A2, and tenascin-X genes. Type IV is characterized by a decreased amount of type III collagen. Types V and VI are characterized by deficiencies in hydroxylase and lysyl oxidase, an important posttranslational modifying enzyme in collagen biosynthesis. Type VII has an amino-terminal procollagen peptidase deficiency. Type IX has abnormal copper metabolism. Type X has nonfunctioning plasma fibronectin.

So the answer to this question is B


Instruction:
In traumatic cases, shock is most likely due to: September 2007
A. Injury to intra abdominal solid organ
B. Head injury
C. Septicemia
D. Cardiac failure

Response:
Ans. A: Injury to intra abdominal solid organ

So the answer to this question is A


Instruction:
Slit-lamp microscopy is indicated in -
A. Corneal Opacities
B. Disease of anterior chamber
C. Lens opacities
D. All of the above

Response:
Structures which can be examined with a slit-lamp are- Lid margin Conjunctiva Cornea Sclera Anterior chamber Iris and pupil Lens Anterior pa of vitreous Reference: A K KHURANA COMPREHENSIVE OPHTHALMOLOGY, Edition 4, page-547

So the answer to this question is D


Instruction:
Maximum protein among pulses is found in-
A. Green gram
B. Red gram
C. Soyabean
D. Black gram

Response:
Ans. is 'c' i.e., Soyabean Amongst Pulses o Highest calories-Soyabeano Highest protein-Soyabeano Highest fat-Soyabeano Highest iron-Soyabeano Highest thiamine-Soyabeano Highest riboflavin-Soyabeano Highest calcium-Horse gramo Highest Niacin-Peas dryo Highest vit C-Bengal gram

So the answer to this question is C


Instruction:
Hea block is seen in:
A. Hypermagnesemia
B. Hypomagnesemia
C. Hypernatremia
D. Hypocalcemia

Response:
Hypermagnesemia slows the hea and leads to hea block Hypomagesemia leads to cardiac arrhythmias like SVT and VT

So the answer to this question is A


Instruction:
Chromosome associated with familial polyposis colon ? -
A. Chromosome 5
B. Chromosome 6
C. Chromosome 11
D. Chromosome 13

Response:
Robbins and Cotran pathologic basis of disease South Asia edition APC is a member of the class of tumor suppressor that function by downregulating growth promoting signalling pathway . Germline loss of function mutation involving the APC which is present in chromosome 5 are associated with familial adenomatous poliposis..

So the answer to this question is A


Instruction:
What is the investigation of choice for interstitial lung disease?
A. Chest X-ray
B. HRCT
C. Gallium - 67 DTPA scan
D. MRI

Response:
Ideally lung biopsy should be done but is not given in the choices. High-resolution computed tomography is superior to the plain chest x-ray for early detection and confirmation of suspected ILD. In addition, HRCT allows better assessment of the extent and distribution of disease. In the appropriate clinical setting HRCT may be sufficiently characteristic to preclude the need for lung biopsy.

So the answer to this question is B


Instruction:
All are sporicidal except:
A. Lysol
B. Glutaraldehyde
C. Ethylene dioxide
D. Formaldehyde

Response:
None

So the answer to this question is A


Instruction:
Flocculo nodular lobe has direct connection with
A. Red nucleus
B. Inferior olivary nucleus
C. Vestibular nucleus
D. Dentate nucleus

Response:
Flocculonodular lobe is also called as vestibulocerebellum. Fibers doesn’t have to project to deep nuclei from here. The efferents from this lobe are sent directly to the vestibular nuclei in the brain stem.

So the answer to this question is C


Instruction:
Conservative management of placenta pre was advocated by _________
A. Cadwell and Moloy
B. Priscilla White
C. Macafee and Johnson
D. Sher and Page

Response:
Conservative management for placenta pre was advocated by Macafee and Johnson. The aim is to continue pregnancy for fetal maturity without compromising maternal health. Prerequisites: Availability of blood transfucion whenever required Facilities for Cesarean section should be available throughout 24 hours. Suitable cases for expectant management for placenta pre: Mother in good health (Hb >/= 10g%, hematocrit >/= 30%) Duration of pregnancy <37 weeks No active vaginal bleeding Fetal well-being is assured (USG) Conduct of expectant treatment: Coicosteroids for fetal lung maturity if <34 week Bed rest Hb, Blood grouping and urine for protein Inspection of vulval pads, fetal surveillance with USG Supplementary hematinics Rh immunoglobin should be given to all Rh-negative (unsensitized) women Other options: Caldwell and Moloy: Classification of Pelvis Sher and Page: Grading of Abruptio placentae Priscilla White: Classification of diabetes in pregnancy. Other named classification/regime in obstetrics Clarke&;s: Classification of hea disease basedd on maternal moality Lytic Cocktail: Treatment of convulsions in eclampsia Ref: DC Dutta&;s Textbook of Obstetrics 8th edition Pgno: 290

So the answer to this question is C


Instruction:
Acetazolamide is administered to a glaucoma patient. Given that this drug inhibits carbonic anhydrase in the renal proximal tubule, which of the following substances will be excreted at a lower rate?
A. Na+
B. H2O
C. HCO3-
D. NH4

Response:
The primary effect of carbonic anhydrase inhibitors such as acetazolamide is to inhibit both H+ secretion and NaHCO3 reabsorption, making the urine alkaline. NH4+ excretion is reduced as a result of the diminished H+ secretion. Carbonic anhydrase inhibitors restrict H+ secretion by inhibiting the intracellular hydration of CO2, a primary source of intracellular H+. The decline in H+ secretion inhibits the Na+-H+ exchange at the luminal membrane of the proximal tubule, which is the primary site of NaHCO3 diuresis. Carbonic anhydrase inhibitors also block the dehydration of H2CO3 formed in the tubular lumen. Chronic doses of such drugs can lead to hyperchloremic acidosis (metabolic acidosis). Ref: McNamara J.O. (2011). Chapter 21. Pharmacotherapy of the Epilepsies. 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 D


Instruction:
Incised-like lacerated wound appears at :
A. Fore head
B. Hand
C. Thorax
D. Abdomen

Response:
A i.e. Forehead

So the answer to this question is A


Instruction:
Interferon-gamma release assays used to diagnose tuberculosis is :
A. T-SPOT.TB
B. B-SPOT.TB
C. NIGRAs
D. C-SPOT.TB

Response:
Two in vitro assays that measure T cell release of IFN with the highly TB-specific antigens ESAT-6 and CFP-10 are available. The T-SPOT.TB is an enzyme-linked immunospot (ELISpot) assay, and the QuantiFERON-TB Gold.

So the answer to this question is A


Instruction:
A 11 month old child presents with complaints of respiratory distress. On examination there is bilateral crepitation and wheezing. Which of the following is the most likely cause –
A. Pneumonia
B. Adenovirus
C. Respiratory synchytial virus
D. Rhinovirus

Response:
The child presenting with respiratory distress and bilateral crepitations and wheezing, we have to suspect brochiolitis.

Differential diagnosis of diseases which are frequently asked

So the answer to this question is C


Instruction:
A district has total population 10 lacs, with under 16 population beings 30%. The prevalence of blindness is 0.8/1000 among under 16 population. Calculate total number of blind among under 16 population in the district -
A. 240
B. 2400
C. 24000
D. 240000

Response:
Ans. is 'a' i.e., 240 o In this question, population of under-16 is 3 lacs (30% of 10 lacs). o Prevalance of blindness is 0.8 every 1000 under-16 population. o So total number of blind among under-16 population = 0.8 x 300000/1000 = 240

So the answer to this question is A


Instruction:
Liquefactive necrosis is typically seen in
A. Ischemic necrosis of the heart
B. Ischemic necrosis of the brain
C. Ischemic necrosis of the intestine
D. Tuberculosis

Response:
Ans. (b) Ischemic necrosis of the brain

So the answer to this question is B


Instruction:
Magistrate inquest is done in case of ?
A. Death by suicide
B. Death by accident
C. Death in psychiatry hospital
D. Death in suscpicious circumstances

Response:
Ans. is 'c' i.e., Death in psychiatry hospital Magistrate's inquest is done under sec 176 CrPC is case of :- Death in prison, reformatories, Borstal school. Death in police costody while under police interrogation. Death due to police firing. Dowry death; i.e. death of a married female less than 30 years of age or death within 7 years of marriage. Exhumation. Death in psychiatry hospital. Custodial rape.

So the answer to this question is C


Instruction:
In winter's WAR line, Amber line represents the:
A. Relative depth of the III molar
B. Point of application of the elevator
C. Bone Level covering the impacted tooth
D. Axial inclination of the impacted tooth relative to II molar

Response:
None

So the answer to this question is C


Instruction:
Complication of TPN is
A. CHF
B. Hypochloremia
C. Metabolic acidosis
D. Leucopenia

Response:
Complications : Related to nutrient deficiency- Hypoglycaemia/hypocalcaemia/hypophosphataemia/hypomagnesaemia(refeeding syndrome) Related to overfeeding: Excess glucose-hyperglycaemia, hyperosmolar dehydration. Hepatic steatosis, hypercapnia, increased sympathetic activity, fluid retention, electrolyte abnormalities. Excess fat-hypercholesterolaemia and formation of lipoprotein X, hyperiglyceridaemia, hypersensitivity reactions. Excess aminoacids: hyperchloraemic metabolic acidosis, hypercalcaemia, aminoacidaemia, uraemia. Related to sepsis: Catheter related sepsis Possible increased predisposition to systemic sepsis Related to line : On inseion : pneumothorax, damage to adjacent aery, air embolism, thoracic duct damage, cardiac perforation or tamponade, pleural effusion, hydromediastinum. Long term use: occlusion, venous thrombosis. Ref: Bailey and love 27th edition Pg no : 288

So the answer to this question is B


Instruction:
A child suffering from acute diarrhoea is brought to the casualty and is diagnosed as having severe dehydration with pH of 7.23. Serum Na-125, Serum K-3, HCO3- 16. The best I.V. fluid of choice is-
A. 3% Saline
B. N/3 Saline +10% dextrose
C. Normal saline
D. N/3 saline+5% dextrose

Response:
Hyponatremia causes severe dehydration Clinical features Patients with hypernatraemia generally have reduced cerebral function, either as a primary problem or as a consequence of the hypernatraemia itself, which results in dehydration of neurons and brain shrinkage. In the presence of an intact thirst mechanism and preserved capacity to obtain and ingest water hypernatraemia may not progress very far. If adequate water is not obtained, dizziness, delirium, weakness and, ultimately, coma and death can result. Management Treatment of hypernatraemia depends on both the rate of development and the underlying cause. If there is reason to think that the condition has developed rapidly, neuronal shrinkage may be acute and relatively rapid correction may be attempted. This can be achieved by infusing an appropriate volume of intravenous fluid (isotonic 5% dextrose or hypotonic 0.45% saline) at an initial rate of 50-70 mL/hr. In older, institutionalised patients, however, it is more likely that the disorder has developed slowly, and extreme caution should be exercised in lowering plasma sodium to avoid the risk of cerebral oedema. Ref Davidson 23rd edition pg 360

So the answer to this question is C


Instruction:
Best bone to assess age between 20-50 yrs.:
A. Skull
B. Ribs
C. Sternum
D. Symphysis pubis

Response:
A i.e. Skull

So the answer to this question is A


Instruction:
Anticoagulant only effective in VIVO:
A. Heparin
B. Coumarin
C. EDTA
D. Sodium citrate

Response:
Ans. b (Coumarin). (Ref. Harrison's internal medicine 16th ed. 688)ANTICOAGULANTSA. Vitamin K antagonists:# The oral anticoagulants act by antagonizing the effects of vitamin K.# They take at least 48 to 72 hours for the anticoagulant effect to develop fully.# Generally, these anticoagulants are used to treat patients with- Deep-vein thrombosis (DVT),- Pulmonary embolism,- Atrial fibrillation, and- Mechanical prosthetic heart valves.# The most important oral anticoagulants are:- Warfarin (Coumarin).- Acenocoumarol- PhenindioneB. Heparin and its derivatives# Heparin is a biological substance, usually made from pig intestines.# It works by activating antithrombin III, which blocks thrombin from clotting blood.# Heparin can be used in vivo (by injection), and also in vitro to prevent blood or plasma clotting in medical devices.# Low molecular weight heparin is a more highly processed product that does not require monitoring of the APTT coagulation parameter (it has more predictable plasma levels) and has fewer side effects.# Fondaparinux is a synthetic sugar composed of the five sugars (pentasaccharide) in heparin that bind to anti- thrombin. It is a smaller molecule than low molecular weight heparin.C. Direct thrombin inhibitors# Include- Argatroban, Lepirudin, and Bivalirudin.- Oral direct thrombin inhibitor are Dabigatran, Rivaroxaban and ximelagatranPD. Anticoagulants outside the body# Apart from heparin, most of the chemicals added work by binding Ca++ ions, preventing the coagulation proteins from using them.- EDTA chemical strongly and irreversibly binds calcium.- Citrate is used for coagulation tests, as well as in blood transfusion bags. It gets rid of the calcium, but not as strongly as EDTA.* It can be in the form of Sodium citrate or ACD.- Oxalate has a similar mechanism to citrate.* It is the anticoagulant used in fluoride tubes.

So the answer to this question is B


Instruction:
Gas shadow in hea and vessels are seen in
A. IUD
B. Still bih
C. Aboion
D. Live IUGR

Response:
Robe sign - the presence of a gas shadow within the hea or the greater vessels, seen in cases of fetal death in utero.It is a rare sign caused by postmoem blood degeneration, usually seen 1-2 days after death; and may be seen as early as 12 hours(Ref: Diagnostic Ultrasound by Carol M. Rumack, Stephanie R. Wilson, J. William Charboneau, and Deborah Levine,4thed, page 1090)

So the answer to this question is A


Instruction:
Advantage of fibrooptic bronchoscopy over rigid bronchoscopy is :
A. Foreign body removal
B. Good view
C. Better airway control
D. In a sick child it can be passed through endotracheal tube

Response:
Ans. is 'b' i.e Good view These days flexible fiberoptic bronchoscopy has replaced rigid bronchoscopy for diagnostic procedures particularly in adults*.It has limited utility in children because of the problems of ventilation.It provides magnification & better illumination.Because of smaller size permits examination of subsegmental bronchi.It is also easy to use in patients with neck or jaw abnormalities where rigid bronchoscopy may be almost impossible technically.Only smaller foreign bodies can be removed. For larger ones rigid bronchoscopy is used.

So the answer to this question is B


Instruction:
Maintenance fluid for a child weighing 10 kg ?
A. 1500 ml/day
B. 1000m1/day
C. 750m1/day
D. 500m1/day

Response:
Ans. is 'b' i.e., 1000m1/day o Up to 10 kg, the requirement is 100 ml/kg/day. o So, Maintenance fluid for a 10 kg child is 1000 ml/kg/day.

So the answer to this question is B


Instruction:
Frequency of breech presentation in pregnancy at term is :
A. 0.10%
B. 1%
C. 3%
D. 10%

Response:
3%

So the answer to this question is C


Instruction:
Major aerial supply to anterior cruciate ligament is from
A. Medial genicular aery
B. Lateral genicular aery
C. Middle genicular aery
D. None of the above

Response:
CRUCIATE LIGAMENTS: Very thick and strong fibrous bands. Direct bonds of union between tibia and fibula. Maintain anteroposterior stability of knee joint. Anterior cruciate ligament:- Begins from anterior pa of intercondylar area of tibia, runs upward, backward and laterally and is attached to the posterior pa of medial surface of lateral condyle of femur. Taut during extension of knee. The anterior cruciate ligament is more commonly damaged than the posterior.It may be injured in violent hyperextension of knee or anterior dislocation of tibia. Posterior cruciate ligament:- Begins from posterior pa of intercondylar area of tibia,runs upwards,forwards and medially and is attached to lateral surface of medial condyle of femur. Taut during flexion of knee. The posterior cruciate ligament is injured in posterior dislocation of tibia. Both these are supplied by middle genicular nerves and vessels. {Reference: BDC 9 E}

So the answer to this question is C


Instruction:
All of the following can cause traumatic Asphyxia, except ?
A. Railway Accident
B. Road Traffic Accident
C. Accidental strangulation
D. Stampede in crowd

Response:
Ans. is 'c' i.e., Accidental strangulation Traumatic asphyxia : It is a form of asphyxia which is caused from respiratory arrest due to mechanical fixation of chest resulting from blunt trauma to chest or external pressure on the chest, abdomen or back which prevents normal respiratory movements of the chest wall. It may be due to :? a) Stampede in crowd (riot crus or human pile death). b) Crush by heavy object e.g. (i) road traffic accident, (ii) railway accident, (iii) oveurn of tractor in farm and (iv) fall of eah, stone etc. in mine or in a building collapse

So the answer to this question is C


Instruction:
Which vaccines are not given in a 8 years old unimmunized child –
A. Pertusis
B. Salk vaccine
C. BCG
D. All of these

Response:
None

So the answer to this question is D


Instruction:
Half-Life of Cobalt-60
A. 30 years
B. 5.2 years
C. 74.5 days
D. 6 hours

Response:
Cobalt-60 – 5.2 years
Cesium-137 – 30 years
Iridum-192 – 74.5days

So the answer to this question is B


Instruction:
Internal splints (fixation devices) are used in all except
A. Compound fractures
B. Multiple fractures
C. Fractures in elderly patients
D. Fracture neck of femur

Response:
A i.e. Compound fracture

So the answer to this question is A


Instruction:
Best sample for DNA karyotyping -
A. Blood
B. Bone marrow
C. Amniotic fluid
D. Chorionic villi

Response:
Ans. is 'a' i.e., Blood* Karyotyping is the study of number and appearance of chromosomes* The first step in karyotyping is sample collection* Sample utilized are# Blood# Bone marrow# Amniotic fluid* Most commonly used sample is blood. Blood is centrifuged. WBCs are separated. White Blood cells are rapidly dividing cells with DNA.* The cells are grown in culture.* During cell cycle, only in M phase of cell cycle, the chromosomes are highly condensed. Hence, metaphase arrest of the cells are mediated using drugs like Colchicine.* The cells are lysed using hypotonic solution.* The chromosomes that are released are stained with Giemsa.* The following features are observed:# Differences in basic number of chromosomes - to detect aneuploidy or trisomy# Differences in absolute sizes of chromosomes - to detect DNA duplication or deletion# Differences in the position of centromeres to detect translocations.# Differences in degree and distribution of heterochromatic regions. Heterochromatin is functionally inactive reghion of a chromosome. It stains darker than euchromatin.* Bone marrow is used as a sample for karyotyping to detect acquired chromosomal aberrations in blood malignancies* Amniotic fluid karyotyping is done only in cases of repeated miscarriages or if copy number variations of chromosomes are highly suspected in the fetus.

So the answer to this question is A


Instruction:
In Japanese Encephalitis, pigs acts as -
A. Amplifier
B. Definitive host
C. Intermediate host
D. Any of the above

Response:
Pigs acts as an amplifier host REF:<\p> ANATHANARAYANAN MICROBIOLOGY NINTH EDITION PAGE.521(epidemiology)

So the answer to this question is A


Instruction:
40 year old male farmer who has livestock was admitted with paroxysms of cough and expectoration. Clinical examination was normal. Chest X-ray showed a large thin walled cavity containing membranes in the lower lobe of the left lung with left pleural effusion. CT scan of the chest revealed a 'water lily' sign. What is the most probable diagnosis?
A. Echinococcus infection
B. Aspergillosis
C. Interstitial lung disease with TB
D. Amebic lung abscess

Response:
Water lily sign although rarely seen is pathognomonic of echinococcosis. Cyst with a freely floating endocyst.

So the answer to this question is A


Instruction:
Golden S is seen in
A. Left upper lobe collapse
B. Pleural Effusion
C. Left lower lobe collapse
D. Bronchogenic carcinoma

Response:
Golden "S" sign is a Reverse "S" sign seen on Frontal Chest Radiograph In patients with Bronchogenic Ca with Right Upper lobe collapse Lower convexity of the Reverse S is due to the mass Upper concavity of the "S" is due to Right UPPER LOBE atelectasis with the upward shifting of the horizontal fissure.

So the answer to this question is D


Instruction:
In-situ DNA nick end labeling can quantitate -
A. Fraction of cells in apoptotic pathways
B. Fraction of cells in S phase
C. p53 gene produce
D. bcr/abl gene

Response:
In Situ Nick-End Labeling An in situ method for detecting areas of DNA which are nicked during APOPTOSIS. Terminal deoxynucleotidyl transferase is used to add labeled dUTP, in a template-independent manner, to the 3 prime OH ends of either single- or double-stranded DNA. The terminal deoxynucleotidyl transferase nick end labeling, or TUNEL, assay labels apoptosis on a single-cell level, making it more sensitive than agarose gel electrophoresis for analysis of DNA FRAGMENTATION. Ref Robbins 9/e p218

So the answer to this question is A


Instruction:
Urine dipstick shows result +3. What's the value of protein appearing in mg/dl in urine is:
A. 100 mg/dL
B. 300 mg/dL
C. 1000 mg/dL
D. 3000 mg/dL

Response:
Ans. b. 300 mg/dLDipstick urinalysisAmount of protein in the urine0Normal1+30 mg/dL2+100 mg/dL3+300 mg/dL4+1000 mg/dL

So the answer to this question is B


Instruction:
Concentration of Local anesthetic agent used for Bier's block is
A. 0.50%
B. 1-2%
C. 4%
D. 5%

Response:
0.5% (Bier's block: Intravenous regional anesthesia)

So the answer to this question is A


Instruction:
Brown color of contusion is due to -
A. Haematodin
B. Reduced hemoglobin
C. Haemosiderin
D. Bilirubin

Response:
Ans. is 'c' i.e., Haemosiderin Color change and age of bruise (Contusion!o After a bruise has appeared, it tends to get smaller from periphery to center and passes through a series of colour changes. These are due to disintegration of RBCs by hemolysis and breakdown of hemoglobin into the pigments haemosiderin, haematodin and bilirubin changes are: -At first: Red (oxyhemoglobin)Few hours to 3 days : blue (reduced hemoglobin)4th day : Bluish-black to brown (haemosiderin)5-6 days: Greenish (haematodin)7-12 days : Yellow (Bilirubin)2 weeks : Normal (absorption of pigment)

So the answer to this question is C


Instruction:
Mineral oils are used in mosquito control measure as -
A. A personal protection method
B. Larvicide
C. Adulticide
D. Space spray

Response:
Mineral oils are used as chemical control against mosquito larvae. The mineral oil application to water is the oldest method to control mosquito. oils used are diesel oil, fuel oil, kerosene oil, and crude oils. Special mosquito larvicidal oils are also available. Oil kills larvae and pupae within a sho time after application. When applied on water oil spreads and forms a thin film, which cuts off the air supply to mosquito larvae and pupae. Since the life cycle of a mosquito is about 8 days, its customary to apply oil once in a week on all breeding places. Other anti-larval measures include source reduction, using Paris green, and synthetic insecticides. Biological control with larvivorous fishes like Gambusia and Lebister. Park's Textbook of Preventive and Social Medicine, 25th Edition, Pg 832

So the answer to this question is B


Instruction:
G protein coupled receptor ?
A. NMDA receptor
B. Steroid
C. M2
D. Insulin receptors

Response:
Ans. is 'c' i.e., M2M2 muscarinic receptors are G-protein coupled receptors and act through cAMP.

So the answer to this question is C


Instruction:
Anuria is defined as urine output less than
A. 4ml/hr
B. 8ml/hr
C. 12ml/hr
D. 16ml/hr

Response:
Anuria is nonpassage of urine, in practice is defined as passage of less than 100milliliters of urine in a day. Anuria is often caused by failure in the function of kidneys. It may also occur because of some severe obstruction like kidney stones or tumours. It may occur with end stage renal disease. It is a more extreme reduction than oliguria (hypouresis), with 100 mL/day being the conventional (albeit slightly arbitrary) cutoff point between the two Ref Harrison20th edition pg 277

So the answer to this question is A


Instruction:
All of the following are features of botulism except:
A. Diplopia
B. Dysphagia
C. Diarrhoea
D. Dysahria

Response:
Botulinum toxin blocks the release of acetylcholine at the neuromuscular junction and parasympathetic postganglionic terminals Dysphagia, dysahria, diplopia, dilated pupils, dry mouth followed by symmetric descending paralysis are the manifestations of botulism. Constipation is seen instead of diarrhea. No fever or sensory abnormalities are seen. Deep tendon reflexes may be normal or decreased. Gag reflex is suppressed. Untreated death occurs due to respiratory paralysis.

So the answer to this question is C


Instruction:
Brushfield spot in iris is seen in -
A. Neurofibromatosis
B. Down syndrome
C. Tuberous sclerosis
D. Toxoplasmosis

Response:
Brushfield spots are seen in Down syndrome. Pedunculated nodules are seen in Neurofibromatosis.

So the answer to this question is B


Instruction:
A person taking tricyclic antidepressants presents with blurred vision and dry mouth. These adverse effects result due to blockade of:
A. M3 muscarinic receptors
B. GABAA receptors
C. H1 histamine receptors
D. 5HT2 receptors

Response:
Ans. (A) M3 muscarinic receptors(Ref: KDT 8/e p486)Anticholinergic adverse effects of TCAs include dry mouth, constipation, blurring of vision, urinary retention, etc.

So the answer to this question is A


Instruction:
Chvostek sign is seen in: March 2013
A. Hypercalcemia
B. Hypothyroidism
C. Insulinoma
D. Diabetes mellitus

Response:
Ans. B i.e. Hypocalcemia Hypocalcemia Idiopathic hypoparathyroidism is associated with: Genetic syndrome of hypoparathyroidism, - Addison's disease and - Mucocutaneous candid iasis Pseudohypoparathyroidism: No deficiency of parathormone, but target organ (bone and kidney) are unresponsive to its action Chvostek's sign: Contraction of facial muscles, elicited by light tapping of the facial nerve ECG changes: Prolongation of QT interval

So the answer to this question is B


Instruction:
Eagle Barrett syndrome, all are true except -
A. Also known as prune belly syndrome
B. Pulmonary hypoplasia
C. Bilateral undescended testes
D. Scapular hypoplasia

Response:
Ans. is 'd' i.e., Scapular hypoplasia Prune Belly or Eagle Barrett syndrome -o Consists of -Extremely lax lower abdominal musculatureDilated urinary tractB/l undescended testeso Comorbidities -Pulmonary' hypoplasiaDislocation/dysplasia of hipPectus excavatum

So the answer to this question is D


Instruction:
Case control study is an example of ?
A. Prospective study
B. Retrospective study
C. Combined retrospective and prospective study
D. Study at one point of time

Response:
Ans. is 'b' i.e., Retrospective study

So the answer to this question is B


Instruction:
A diabetic patient in hypoglycemia does not regain consciousness despite blood glucose restoration to normal. Which one of the following is the not likely condition ?
A. Cerebral edema
B. Alcohol intoxication
C. Post-ictal state
D. Cerebral haemorrhage

Response:
Alcohol consumption causes an increase in insulin secretion, which leads to low blood sugar (otherwise known as hypoglycaemia). This causes light headedness and fatigue, and is also responsible for a host of longer-term alcohol-related health problems . REf - pubmed.com

So the answer to this question is B


Instruction:
A patient Amit Kumar is suffering from atherosclerosis. Which of the following lowing is the most beneficial drug for prevention of stroke in this patient?
A. Aspirin
B. Warfarin
C. Low dose subcutaneous herparin
D. Digoxin

Response:
Ref: KDT 6/e p610,611 Antiplatelet drugs are used for the prophylaxis of aerial thrombotic conditions like stroke and MI. Atrial fibrillation increase the risk of thromboembolism and can result in stroke.

So the answer to this question is A


Instruction:
The following bacteria doesn't produce any spores
A. Bacillus anthracis
B. Clostridium
C. Pseudomonas
D. Bacillus subtilis

Response:
Pseudomonas doesn't produce spores. Bacillus and Clostridium produce spores Ref: Textbook of Microbiology Baveja 5th ed Pg 21

So the answer to this question is C


Instruction:
Which of the following antiepileptic agents acts on the GABAergic system to decrease the uptake of GABA into neurons and glial cells?
A. Vigabatrin
B. Progabide
C. Gabapentin
D. Tiagabine

Response:
It potentiates GABA mediated neuronal inhibition by depressing GABA transpoer GAT 1 which removes synaptically released GABA into neurons and glial cells REF: KD TRIPATHI 8TH ED. page 421

So the answer to this question is D


Instruction:
All action occur in mitosis EXCEPT
A. Cytokinesis
B. Chromatids separate
C. DNA replication
D. The kinetochore becomes evident

Response:
DNA replication or DNA synthesis occurs during the S phase of the cell cycle. G1 S, and G2 make up interphase. Mitosis completes the cell cycle and is composed of prophase, metaphase, anaphase, and telophase. Cytokinesis is the division of the cell cytoplasm that occurs in telophase and is orchestrated by the microfilaments. Chromatids separate and move toward the poles of the cell in anaphase. The nuclear membrane and nucleolus disappear during metaphase.

So the answer to this question is C


Instruction:
On the basis of histologic features, Internal thoracic artery is
A. Medium sized arksey
B. Small sized artery
C. Large sized artery
D. Capillary

Response:
Internal thoracic artery is a medium sized artery with histological features of Large sized artery

So the answer to this question is C


Instruction:
Risk of preterm delivery is increased if cervical length is:
A. 2.5 cm
B. 3.0 cm
C. 3.5 cm
D. 4.0 cm

Response:
None

So the answer to this question is A


Instruction:
The following is a feature of Korsakoff’s psychosis
A. Confabulation
B. Encephalopathy
C. Opthalmoplegia
D. Hallucination

Response:
Ref. Oxford textbook of Psychiatry. Page. 28

Anterograde amnesia is also observed in patients with Korsakoff syndrome.

Korsakoff syndrome is seen mainly in alcoholics who have a thiamine deficiency and often follows an acute presentation of Wernicke encephalopathy.
Wernicke encephalopathy presents with ocular palsies, confusion, and gait ataxia and is also related to a thiamine deficiency.
In Wernicke-Korsakoff syndrome, lesions are always found in the mammillary bodies and the dorsomedial nuclei of the thalamus.
In addition to exhibiting an anterograde amnesia, Korsakoff patients also present with retrograde amnesia.
These patients confabulate, making up stories to replace past memories they can no longer retrieve.

So the answer to this question is A


Instruction:
Which of the following is not a component of mackler's triad for esophageal perforation
A. Chest pain
B. Vomiting
C. Subcutaneous emphysema
D. Aspirational pneumonitis

Response:
Mackler’s triad consists of chest pain, vomiting and subcutaneous emphysema.

So the answer to this question is D


Instruction:
In vitro test for virulence of pneumococcus:
A. Intra peritoneal inoculation in mice
B. Rat tail inoculation
C. Mouse foot pad inoculation
D. Armadillo inoculation

Response:
Ans: a (Intra peritoneal inoculation in mice) RefAnanthanarayan, ed, p. 220When pneumococci are scanty with negative cultures, isolation can be done by intra peritoneal inoculation in mice for virulent strains. Inoculated mice will die within 1 -3 days and pneumococci may be isolated from peritoneal fluid or heart blood. This will be negative with avirulent strains e.g., type 14.Other tests employed are:1) Gram stain - gram positive, lanceolate diplococcic2) Sputum culture in blood agar3) Blood culture - isolation from blood indicates bad prognosis4) CSF - Gram staining and cultureDemonstration of SSS by precipitation5) Capsular polysaccharide - Quellung reaction, counter immune electrophoresis

So the answer to this question is A


Instruction:
Type I MEN involves all, except -
A. Pancreas
B. Adrenal
C. Pituitary-'
D. Parathyroid

Response:
Ans. is 'b' i.e., Adrenal Characteristic triad of MEN IParathyroid tumors (Most common)Pancreatic endocrine tumors (Second most common)Pituitary tumors (Third most common)Hyperparathyroidism is the mostCommon manifestation of MEN I.-- HarrisonNeoplasia of the pancreatic islet is the second most common manifestation of MEN1.Pituitary tumors are third most common tumors of MEN / and occur in more than half of patients with MEN I.Relative frequency of tumors causing hyperparathyrodism :Relative frequency of endocrine tumors of pancreatic islets:Relative frequency ofpituitary tumors:o Parathyroid hyperplasia is the most common cause of hyperparathyroidism in MEN Io Pancreatic polypeptide (75-82%)o Prolactinomas are the most common pituitary tumors in MEN Io Gastrin (60%) (Gastrinomas/ZES)o Insulin (25-30%) (Insulinomas)o Acromegaly due to excessive growth hormone (GH) production is the second most common pituitary tumor in MEN Io Parathyroid adenomas are less common than parathyroid hyperplasia and tend to occur in older patient or those with long standing disease.o Glucagon (5-10%) (Glucagonoma)o VIP (3-5%) (VIPomas)o Somatostatin (1-5%) (Somatostatinoma)o Cushing s syndrome Other tumors in MEN Io Pheochromocytomaso Carcinoid tumours (foregut type : derived from lung; thymus, stomach or duodenum).o Subcutaneous and visceral lipomas.o Cutaneous leiomyomas.o Skin angiofibromas or collagenomas.

So the answer to this question is B


Instruction:
In Gaucher's diseases, there is deficiency of ?
A. Glucocerebrosidase
B. Glucokinase
C. Sphingomyelinase
D. G-6PD

Response:
Ans. is 'a' i.e., Glucocerebrosidase

So the answer to this question is A


Instruction:
True about point source epidemic is?
A. Occurs in more than 1 incubation period
B. Occurs in one incubation period
C. The exposure is continuous
D. Epidemic curve falls very slowly

Response:
Common source single exposure epidemics is also known as point source epidemic. The exposure to the disease agent is brief and essentially simultaneous, the resultant cases all develop within one incubation period of the disease. Ref: 25th edition, Park's Textbook of Preventive and Social Medicine, Page no. 72

So the answer to this question is B


Instruction:
A study of colonic polyps is performed. Malignant cells localized to the polyp are compared to those from polyps showing invasion of the stalk. Molecular analysis shows up-regulation of certain molecules in the invasive malignant cells. Invasive lesions are more likely to exhibit lymphatic metastases. Which of the following markers is most likely to have increased expression in the invasive malignant epithelial cells?
A. BCL2
B. CD44
C. EGFR
D. RAS

Response:
The malignant transformation includes many genetic changes, including those rendering the malignant cells capable of invasion and metastases. CD44 plays a role in cell adhesion and enables malignant cells to metastasize. Solid tumors can express CD44 to enhance their spread to lymph nodes and other metastatic sites. Though such properties may have been present with the initial clone of malignant cells, the growth of cancer increases the number of cells and the risk for spread. BCL2 plays a role in apoptosis. Growth factor receptor expression may make malignant cells susceptible to environmental influences, such as hormones, that drive growth. RAS gene mutations are present in many cancers and lead to loss of growth control. Vimentin is best known as a protein expressed in mesenchymal neoplasms, such as sarcomas, but it is also up-regulated in EMT.

So the answer to this question is B


Instruction:
Low glycemic index food is:
A. Easily digestible
B. Increases plasma glucose
C. Has slower absorption
D. Increase glycogen deposits

Response:
GLYCEMIC INDEX -glycemic index which measures the time course of postprandial glucose concentrations from a graph* Glycemic index of complex carbohydrate is low because of slow digestion and slow absorption* Glycemic index of carbohydrate is low when its combine with fat , protein and fiber * Foods that have a low glycemic index are considered to be more beneficial since they cause less fluctuation in insulin secretion

So the answer to this question is C


Instruction:
A boy gets hit by a tennis ball in the eye following which he has complaints of decreased vision. Which of the following tells that blunt injury is due to the ball?
A. Optic neuritis
B. Pars planitis
C. Vitreous base detachment
D. Equatorial edema

Response:
Amongst the given options, only vitreous base detachment is pathognomonic for blunt trauma and is associated with retinal detachment.

"Blunt trauma to eye can cause linear tears of retina along the anterior border or posterior border of vitreous base. Such tears are most common superonasally and inferotemporally and may be accompanied by an avulsed vitreous base, which is pathognomonic of traumatic retinal detachment".  —Atlas of clinical ophthalmology
"Sometimes the retina tear along both the anterior and posterior borders of the vitreous base, the base itself is avulsed and may having like a pigmented loop in the vitreous cavity with its underlying strip of attached retina superonasal avulson of vitreous base is pathognomonic of traumatic retinal detachment". — Harley's pediatric ophthalmology

So the answer to this question is C


Instruction:
Employer's contribution to Employees State Insurance
A. 3.25
B. 3.5
C. 3.75
D. 3.2

Response:
Ref:According to ESI Act amendment 2019

So the answer to this question is A


Instruction:
The laboratory evidence of flare in SLE is:
A. Increase in titre of anti-dsDNA
B. Increase in titre of ANA
C. Increase in anti-Sm
D. Elevation of ESR

Response:
SLE FLARE: LAB RESULTS Increases in quantities of anti-dsDNA herald a flare, paicularly of nephritis or vasculitis It may be associated with declining levels of C3 or C4 complement. Antibodies to Sm is specific for SLE and assist in diagnosis; anti-Sm antibodies do not correlate with disease activity or clinical manifestations. Ref: Harrison, E-18, P-2735

So the answer to this question is A


Instruction:
True statement regarding migraine in pregnancy is:
A. Ketorolac is preferred
B. NSAIDs are first line drugs
C. Improved symptoms with pregnancy
D. Sumatriptan preferred over chlorpromazine

Response:
Ans: (b) NSAIDs are first line drugsRef: Harrison's Principles of Internal Medicine, 19th edition, Page 2592Treatment of Acute MigraineNSAIDsNaproxen, lbuprofen,Tolfenamic acid Diclofenac, Acetominophen5-HT1 Receptor AgonistsOral: Ergotamine, Naratriptan, Rizatriptan, Sumatriptan, Frovatriptan, Almotriptan, Eletriptan, ZolmitriptanNasal: Sumatriptan, Zolmitriptan Parenteral: Dihydroergotamine, SumatriptanDopamine Receptor AntagonistsOral: Metoclopramide, ProchlorperazineParenteral: Chlorpromazine, Metoclopramide, ProchlorperazineOthersNasal: Butorphanol Parenteral: OpioidsMigraine During PregnancyImprovement in 'migraine without aura'Worsening of 'migraine with aura'Ergot derivatives are contraindicated due to possibility of uterine hypertonicity and vascular disruption leading on to miscarriageSafety of Triptans during pregnancy is not fully confirmed. Can be used only if other drugs are not effectiveTreatmentFirst line: NSAIDs and paracetamolTriptans are avoidedProphylaxis: Magnesium, Metoprolol

So the answer to this question is B


Instruction:
Therapeutic uses of Prostaglandin El include all of the following except
A. Medical termination of pregnancy
B. Impotence
C. Primary pulmonary hypertension
D. Maintenance of patent ductus arteriosus (PDA)

Response:
Ans. is 'c' primary pulmonary hypertensionRef: K.D.T. 5th/e page 164, 165 (4th/ep 219), Goodman Gilman 10th/ep679, Dutta obstetrics 5th/e, p187)Uses of PGE1 (Misoprost)Gastric ulcer-PGE1 reduces the basal acid secretion by 90% and stimulated acid secretion by 80% so they heal gastric ulcer ulcer healing rates are comparable to cimetidineImpotence-PGE1 may be used in the treatment of impotence.Intracavernous injection of PGE1 causes partial erection in impotent patients.PGE1 is more effective than papaverine. Priapism occurring sometimes with papaverine/ phentolamine inj. has not been noted with PGE1.Maintenance of patent ductus arteriosus-The ductus arteriosus in neonates is highly sensitive to vasodilation by PGE1.Patency of ductus arteriosus is necessary in some patients with congenital heart disease till surgery is undertaken PGE1 has proved to be highly effective in such patients.Medical Termination of PregnancyUse of prostaglandins in primary pulmonary hypertensionPrimary pulmonary hypertension is a rare idiopathic disease mainly observed in young adults. It usually leads to right heart failure. Lung or lung-heart transplantation has been the treatment previouslyLong term therapy with PGI2 (epoprostenol) offers the following benefitsDelayed or avoided the need for transplant Marked improvement in symptomsPGE1 has no role in primary pulmonary hypertension

So the answer to this question is C


Instruction:
Following genetic counselling in a family for Familial polyposis coli (FPC) next screening test is
A. Flexible sigmoidoscopy
B. Colonoscopy
C. Occult blood in stools
D. APC gene

Response:
Ans. is 'd' i.e. APC gene Schwartz writes - "Flexible sigmoidoscopy of first-degree relatives of FAP (Familial adenomatous polyposis) patients beginning at age 10 to 15 years has been the traditional mainstay of screening. Today following genetic counselling, APC gene testing may be used to screen family members providing an APC mutation has been identified in a family member."Familial adenomatous polyposis (FAP) is a dominantly inherited colon cancer syndrome due to germline mutations in the adenomatous polyposis coli (APC) tumor suppressor gene on chromosome 5.Patients with this syndrome develop hundreds to thousands of adenomas in the colon which if left untreated will eventually develop into colon cancer.Each of the FAP adenomas has lost the normal remaining allele of APC gene but has not yet accumulated the required additional mutations to generate fully malignant cells. In due course of time many of these adenomas acquire further genetic abnormalities and develop into fully malignant cancers. APC gene is thus considered to be a gatekeeper for colon tumorigenesis.The detection of APC gene mutation in family members of a FAP patient helps in making a definitive diagnosis before the development of polyps.So APC gene testing is used to screen family members

So the answer to this question is D


Instruction:
Cfitoris develops from
A. Urogenital Sinus
B. Labio Scrotal Swelling
C. Genital folds
D. Genital tubercle

Response:
Genital tubercle

So the answer to this question is D


Instruction:
The resting membrane potential of a cell:
A. Is dependent on the permeability of the cell membrane to K+ being greater to Na+
B. Falls to zero if Na+/K+ ATPase in membrane is inhibited
C. Is equal to the equilibrium potential for K+
D. Is equal to the equilibrium potential of Na+

Response:
Is dependent on the permeability of the cell membrane to K+ being greater to Na+ REF: Ganong, 22nd edition, BRS physiology, 4th edition page 11 Ionic Basis of nerve Resting Membrane Potential: Resting membrane potential is potential difference across the cell membrane in millivolts. Which is by convention -70 Mv Resting membrane potential is established by diffusion potential that results from concentration differences of permeable ions. And each permeable ion attempts to derive the membrane potential towards its equilibrium potential. Resting membrane potential (-70 mV) is close to equilibrium potential of K+ and C1-- (both -85 mV) and far away from the equilibrium potential of Na+ ( +65 mV). That means at rest the nerve membrane is more permeable to K+ than Na+ Na+ is actively transpoed out of neurons and other cells and K+ is actively transpoed into cells, but because of K+ permeability at rest is greater than Na+ permeability. Therefore, K+ channels maintain the resting membrane potential. Ionic Basis of nerve Action Potential: Depolarization causes rapid opening of the activation gates of Na+ channels and thus Na+ conductance promptly increases. Thus the membrane potential is driven towards the equilibrium potential of Na+ ion (+65 mV) Thus the rapid depolarization during upstroke is due to inward Na+ current Depolarization slowly opens K+ channels and increases K+ conductance to even higher levels than at rest. The combined effect of closing Na+ channels and greater opening of K+ channels make the K+ conductance higher than the Na+ conductance and membrane potential is repolarized. Thus repolarization is caused by outward K+ current. The K+ conductance remains higher than at rest for some time after closure of Na+ channel and the membrane potential is driven more closer to K+ equilibrium potential than at rest. This is called as undershoot.

So the answer to this question is A


Instruction:
A10-year-old male child complains of pain and swelling in the left foot. He gives a history of injury while playing. Migration of leukocytes to the site of injury is mediated by
A. Cytokines
B. Histamine
C. Chemokines
D. Prostaglandins

Response:
LEUKOCYTE RECRUITMENT
• The journey of leukocytes from the vessel lumen to the tissue is a multistep process that is mediated and controlled by adhesion molecules and cytokines called chemokines
LEUKOCYTE ADHESION
• Extravasation is the sequence of events in the journey of leukocytes from vessel lumen to interstitial tissue which include the following steps
> Margination - leukocytes assume a peripheral portion of lumen along endothelial surface
> Rolling - leukocytes adhere transiently, detach and bind again and finally adhere firmly
> Rolling is mediated by Selectins (P, L and E-selectins)
> Firm adhesion is mediated by Integrins (LFA-1. MAC-1, VLA-4, α4β7)
> Pavement - endothelium virtually lined by leukocytes
LEUKOCYTE MIGRATION
• Diapedesis or transmigration
> Leukocyte migration through endothelium
> Occurs mainly in post capillary venules
> Adhesion molecules involved - CD31 or PECAM-1 (platelet endothelial cell adhesion molecule)
CHEMOTAXIS
• Leukocyte migration toward the site of injury
• Locomotion oriented along a chemical gradient
• Chemo-attractants
> Exogenous - bacterial products (most common)
> Endogenous
■ Cytokines, particularly - IL-8
■ Complement, particularly - C5a
■ Arachidonic acid metabolites particularly - LTB4
• Neutrophils predominate in the inflammatory infiltrate during the first 6 to 24 hours
• Neutrophils are replaced by monocytes in 24 to 48 hours
• In Pseudomonas infection - cellular infiltrate is dominated by neutrophils for several days
• In viral infections, lymphocytes may be the first cells to arrive

So the answer to this question is C


Instruction:
All are true about human mitochondrial DNA except
A. Circular
B. AGA is stop codon
C. Low mutation rate
D. Very few untranslated sequences

Response:
Major Features of Human Mitochondrial DNA Is circular, double-stranded, and composed of heavy (H) and light (L) chains or strands Encodes 13 protein subunits of the respiratory chain (of a total of about 67) Encodes large (16S) and small (12S) mt ribosomal RNAs Encodes 22 mt tRNA molecules Genetic code differs slightly from the standard code UGA (standard stop codon) is read as Trp AGA and AGG (standard codons for Arg) are read as stop codons Contains very few untranslated sequences High mutation rate (5-10 times that of nuclear DNA) Comparisons of mtDNA sequences provide evidence about evolutionary origins of primates and other species Harper30e pg: 378

So the answer to this question is C


Instruction:
In which order the airway of the neonates, who require suctioning, be suctioned during neonatal resuscitation?
A. Nose followed by mouth
B. Mouth followed by nose
C. Pharynx followed by oral cavity followed by trachea
D. Nose followed by Trachea

Response:
The mouth is suctioned before the nose ('M' before 'N') To ensure that the infant does not aspirate, if he gasps when the nose is suctioned.

So the answer to this question is B


Instruction:
The DCCT (Diabetes Control & Complication Trial) provided definitive proof that reduction in chronic hyperglycemia helps to improve:
A. Micro vascular complications of Type I DM
B. Macro vascular complications of Type I DM
C. Micro vascular complications of Type II DM
D. Macro vascular complications of Type II DM

Response:
None

So the answer to this question is A


Instruction:
The drug which is not suitable for patients with acute porphyria for intravenous induction is:
A. Thiopentone sodium
B. Propofol
C. Midazolam
D. Etomidate

Response:
Ans. is 'a' i.e., Thiopentone sodium It has been repeated so many times that barbiturates can induce fatal attacks of porphyria in patients with acute intermittent or variegate porphyria and are contraindicated in such patients.Also know

So the answer to this question is A


Instruction:
Swarming growth on culture is characteristic of which Gram-negative organism
A. Clostridium welchii
B. Clostridium tetani
C. Bacillus cereus
D. Proteus mirabilis

Response:
1. Swarming - Proteus has an ability to swarm (or spread) on solid media. They spread on the surface of the plate in successive waves to form a thin filmy layer in concentric circles. This is known as swarming. 2. Organism which swarm: Proteus, Clostridium tetani, V. parahemolyticus, and Serratia. Clostridium welchii - non motile Clostridium tetani - Gram positive Bacillus cereus- motile, gram positive rod Proteus mirabilis - peritrichous flagella

So the answer to this question is D


Instruction:
The thickest nerve of the body is?
A. Radial
B. Median
C. Sciatic
D. Axillary

Response:
Sciatic nerve (L4-S3) is the largest peripheral nerve in the body.It is the longest and widest single nerve in the human body, going from the top of the leg to the foot on the posterior aspect The sciatic nerve is comprised of the tibial and common peroneal nerve and exits the pelvis inferior to the piriformis muscle, between the ischial tuberosity and the greater trochanter of the femur.

So the answer to this question is C


Instruction:
A 26 year old female complaining of scratchy croaky voice. Which of the following is true:
A. The best treatment is botulinum in cricoarytenoid
B. Type I thyroplasty can be used as treatment
C. Adductor dysphonia is the cause
D. Abductor dysphonia is the cause

Response:
Scratchy croaky is a high pitched sound arising due to rubbing of vocal cords in adductor dysphonias. The best treatment here will be percutaneous botulinum injection in the thyroarytenoid muscle.

So the answer to this question is C


Instruction:
Bater syndrome is mutation in
A. Na-k-cl co transpoer
B. Na-Cl channel
C. Renin
D. ENaC protein

Response:
Bater syndrome is a group of similar rare conditions that affect the kidneys. It&;s genetic, which means it&;s caused by a problem with a gene. If you have it, too much salt and calcium leave your body when you pee. It also may cause low levels of potassium and high levels of acid in the blood. Reference: GHAI Essential pediatrics, 8th edition

So the answer to this question is A


Instruction:
Wahin-Finkeldey giant cells are seen in -
A. Measles
B. Rubella
C. Influenza
D. Rickettsial pox

Response:
Ans. is 'a' i.e., Measles o Wahin-Finkeldey cells are multinucleated giant cells. o These cells are pathognomic of measles and are found in lymphoid organs, lung and sputum.

So the answer to this question is A


Instruction:
Which dietary vitamin is required for transaminase reactions?
A. Thiamine
B. Folic acid
C. Pyridoxine
D. Cyanocobalamin

Response:
Pyridoxal phosphate paicipates in transamination by cycling between the pyridoxal and pyridoxamine forms. Thiamine is a cofactor for transketolase and the -ketoacid dehydrogenases. Folate functions in single carbon transfer reactions. Vitamin B12 functions in methionine synthase and methylmalonyl CoA mutase.

So the answer to this question is C


Instruction:
Posthumous child is one born to a:
A. After mother death
B. After father death
C. Dead born child
D. Both parents death

Response:
Ans. (B). After father death(Ref: Essentials of forensic medicine & toxicology, KSN Reddy, 33rd edition; page 396)A posthumous birth is a birth of a child after the death of a biological parent.A person born in these circumstances is called a posthumous child or a posthumously born person.

So the answer to this question is B


Instruction:
An IQ between 50 & 70 would be classified as what kind mental retardation -
A. Mild
B. Moderate
C. Severe
D. Borderline

Response:
Ans. is 'a' i.e., Mild

So the answer to this question is A


Instruction:
Vagabond&;s disease is also known as
A. Pediculosis pubis
B. Peduculosis capitis
C. Peduculosis corporis
D. Tinea capitis

Response:
pediculosis corporals/body louse: causative organism: pediculus humanus var corporis. Body lice live in clothing and beds and feed on human blood seen in poor and in those of low intelligence infection is transmitted by clothes and beds. clinical features: itching, erythematous macules, uicaria wheals, excoriated papules parallel linear scratch marks. parellel liner scratch marks with post inflammatory hyperpigmentation over the upper back was called vagabonds disease treatment: proper washing and drying of clothes and bedding dusting garments with gamma benzene he a chloride r malathion body louse acts as a vector for the following diseases epidmic typhus-rickettsia prowazekii relapsing fever--baonella recurentis trench fever- baonella quintana iadvl textbook of dermatology page 412

So the answer to this question is C


Instruction:
True about Wilson's disease is ?
A. Increased serum ceruloplasmin
B. Decreased liver copper
C. Increased urinary copper excretion
D. Decreased urine copper excretion

Response:
None

So the answer to this question is C


Instruction:
The most impoant stimulant of respiratory centre is:
A. Alkalosis
B. Decrease PCo2
C. Increase PCo2
D. Decrease Pao2

Response:
Response to Co2, PaCO2 is most impoant factor in control of ventilation under normal conditions. PCo2 is most impoant input regulating magnitude of ventilation under resting conditions. Changes in alveolar ventilation have immediate, pronounced effect on aerial Pco2 (unlike Po2). Even slight alterations from normal Pco2 induce significant reflex. Increased Pco2 increases ventilation. Blood Brain Barrier is permeable to Co2, so increased aerial Pco2 increases brain ECF Pco2 and , it stimulates central chemoreceptors which increases ventilation by stimulating respiratory centers. Regulation of respiration: The rhythmic discharges from the brain that produce spontaneous respiration are regulated by alterations in aerial Po2, PCo2, and H+ concentration. There are numerous factors that affect and control ventilation. Chemical control Co2 ( CSF and brain interstitial fluid H+ concentration) o2 ( carotid and aoic bodies) H+ ( carotid and aoic bodies) Nonchemical control Vagal afferents from receptors in the airways and lungs Afferents from the pons, hypothalamus, and limbic system Afferents from proprioceptors Afferents from baroreceptors: aerial, atrial, ventricular, pulmonary Preipheral (Carotid & Aoic) Chemoreceptors: The carotid bodies are located at the bifurcation of common carotid aery. They send afferents in the carotid sinus nerve to the glossopharyngeal nerve (IX) the aoic bodies are located between the arch of aoa and pulmonary aery afferents ascend in the recurrent laryngeal nerves to the vagus (X). Stimulation results from a decrease in carotid and aoic body tissue PO2. Central Chemoreceptors: situated near (beneath) the ventral surface of the medulla, near the origins of the vagi and glossopharyngeal nerves these are anatomically separate from the respiratory centres, and are bathed in brain ECF. This is impermeable to both H+ and HCO3 -, however CO2 diffuses readily and decreases pH of CSF within a few minutes, which subsequently increases ventilation. Ref: A Concise Textbook Of Physiology, By S. And Kutty, K.M., Page 101.

So the answer to this question is D


Instruction:
Hallucination is a disorder of -
A. Thought
B. Perception
C. Memory
D. Intelligence

Response:
Altered perception :- Sensory distortion (micropsia, hyperacusis), Illusion.
False perception :- Hallucination.

So the answer to this question is B


Instruction:
Glucuronic acid and Iduronic acid are:
A. Anomers
B. Enantiomers
C. Functional isomers
D. Epimers

Response:
option d : Glucuronic acid and Iduronic acid are Epimers Epimers are stereoisomers that differ around the configuration of a single asymmetric carbon other than the anomeric carbon. Epimers of glucose can be remembered as MAGI. Mannose -C2 epimer, Allose -C3 epimer, Galactose -C4 epimer Idose -C5 epimer.

So the answer to this question is D


Instruction:
Gastric lavage is contraindicated in poisoning with -
A. Sulphuric acid
B. Organophosphorus compounds
C. Arsenic
D. Dhatura

Response:
Gastric Lavage * Indications -- Conscious patient -- Ingestion of poison within 1 to 2 hour. * Contraindications -- Coma/convulsion/unconscious patient -- Impaired gag reflex -- Corrosive/volatile poison ingestion -- Marked hypothermia -- esophageal varices -- Significant electrolyte imbalance. * Complications -- Aspiration pneumonia -- Mallory Weiss tears -- Perforation of esophagus/stomach -- Laryngospasm -- Hypothermia -- Electrolyte imbalance * Procedure -- Protect airway -- Place patient on lateral position -- Tube should be passed orally -- Lubricate the inseing end of tube -- Use mouth gag -- The position of tube in stomach should be confirmed -- Lavage is carried out with saline or water. * Precautions -- Never embark on procedure as routine -- Do not do in non-toxic agent ingestion -- Obtain consent and explain procedure * Agents used -- Ewald tube -- Lavacuator -- Ryle's tube (in children) -- Boas tube REF:THE SYNOPSIS OF FORENSIC MEDICINE AND TOXICOLOGY:K S NARAYANA REDDY;PAGE NO 287

So the answer to this question is A


Instruction:
Pain of musculoskeletal origin is all except
A. MPDS
B. TMJ arthritis
C. Pos traumatic pain
D. Stylohyoid eagle’s syndrome

Response:
Pain of musulo skeletal origin

MPDS
TMJ arthritis
Stylohyoid eagles’s syndrome

So the answer to this question is C


Instruction:
All are seen in Marasmus except ?
A. Hepatomegaly
B. Muscle wasting
C. Voracious appetite
D. Weight loss

Response:
Ans. is 'a' i.e., Hepatomegaly Hepatomegaly is seen in kwashiorkor (not in marasmus). Marasmus o It is characterized by gross wasting of muscle and subcutaneous tissues resulting in emaciation and marked stunting. o There is no edema. o Body weight is less then 60% of expected. o Fat in adipose tissues is severely depleted. However the buccal pad of fat is preserved till the malnutrition becomes extreme because a higher propoion of saturated fatty acids is stored there and the saturated fat is the last to be depleted. o Skin is dry, scaly and inelastic with wrinkles. o The hair is hypopigmented. o Abdomen is distended due to wasting and hypotonia of abdominal wall muscles. o The child is ale but irritable. o Child may show voracious appetite.

So the answer to this question is A


Instruction:
The receptor through which M-tropic HIVstrains bind -
A. CCR5
B. CXR4
C. CXCR5
D. Any of the above

Response:
Ans. is 'a' i.e., CCR5 HIV tropism HIV tropism refers to the cell type that the human immunodeficiency virus (HIV) infects and replicates in. HIV tropism of a patient's virus is measured by the Trofile assay. HIV can infect a variety of cells such as CD4+ helper T- cells and macrophages that express the CD4 molecule on their surface. HIV-1 entry to macrophages and T helper cells is mediated not only through interaction of the virion envelope glycoproteins (gp120) with the CD4 molecule on the target cells but also with its chemokine coreceptors. Macrophage (M-tropic) strains of HIV-1, or non-syncitia-inducing strains (NSI) use the beta-chemokine receptor CCR5 for entry and are thus able to replicate in macrophages and CD4+ T-cells. These strains are now called R5 viruses. The normal ligands for this receptor, RANTES, macrophage inflammatory protein (MIP)-1-beta and MIP- I -alpha, are able to suppress HIV-1 infection in vitro. This CCR5 coreceptor is used by almost all primary HIV-1 isolates of viral genetic subtype. T-tropic isolates, or syncitia-inducing (SI) strains replicate in primary CD4+ T-cells as well as in macrophages and use the alpha-chemokine receptor, CXCR4, for entry. These strains are now called X4 viruses. The alpha-chemokine, SDF-1, a ligand for CXCR4, suppresses replication of T-tropic HIV-1 isolates. It does this by down regulating the expression of CXCR4 on the surface of these cells. Viruses that use only the CCR5 receptor are termed R5, those that use CXCR4 are termed X4, and those that use both X4R5. However, the use of coreceptor alone does not explain viral tropism, as not all R5 viruses are able to use CCR5 on macrophages for a productive infection.

So the answer to this question is A


Instruction:
A 35 - year old male who had chronic duodenal ulcer for the last six years presents with worsening of symtomps loss of periodicity of symptoms, pain on rising in the morning sense of epigastric bloating and post - prandial vomiting. The most likely cause of the worsening of his symptoms is the development of -
A. Posterior penetration
B. Gastric outlet obstruction
C. Carcinoma
D. Pancreatitis

Response:
None

So the answer to this question is B


Instruction:
One of your obstetric patients presents to the office at 25 weeks complaining of severe left calf pain and swelling. The area of concern is slightly edematous, but no erythema is apparent. The patient demonstrates a positive Homan sign, and you are concerned that she may have a deep vein thrombosis. Which of the following diagnostic modalities do you order?
A. MRI
B. Computed tomographic scanning
C. Venography
D. Real-time ultrasonography

Response:
Noninvasive modalities are currently the preferred tests for diagnosing venous thromboembolism. Venography is still the gold standard, but it is not commonly used because it is cumbersome to perform and expensive and has serious complications. Real-time ultrasonography or color Doppler ultrasound is the procedure of choice to detect proximal deep vein thrombosis. MRI and CT scanning are used in specific cases when ultrasound findings are equivocal.

So the answer to this question is D


Instruction:
The first product of glycogenolysis is:
A. Glucose-6-phosphate
B. Glucose- 1,6 diphosphate
C. Glucose-1-phosphate
D. Fructose-1-phosphate

Response:
Glycogenolysis, the breakdown of glycogen to release glucose, is an impoant pathway in energy use and reactions that utilize carbohydrates. Glycogenolysis involves three steps. First, the repetitive removal of glucose residues by phosphorylase breaking bonds between carbons 1 and 4, producing a glucose-1-phosphate molecule. However, this process stops when four glucose residues are left before the carbons 1 and 6 branch points. Second, a glucan transferase enzyme moves three residues to the linear pa of the glycogen chain, leaving only the final branched glucose residue. Third, a debranching enzyme cleaves the carbons 1 and 6 bond. A final enzyme conves the resulting glucose-1-phosphate molecules into glucose-6-phosphate for entry into glycolysis. Ref: Janson L.W., Tischler M.E. (2012). Chapter 6. Carbohydrate Metabolism. In L.W. Janson, M.E. Tischler (Eds), The Big Picture: Medical Biochemistry.

So the answer to this question is C


Instruction:
Most common cancer arising from a burns scar is:
A. Adeno carcinoma
B. Fibrosarcoma
C. Squamous carcinoma
D. Adeno squamous carcinoma

Response:
Ans: C (Squamous carcinoma) Ref: Bailey & Love's Short Practice of Surgery, 25th Edition 611Explanation:Squamous carcinoma of SkinSecond most common form of skin cancerTwice more common in menRisk factorsUV rays exposureChronic inflammationChronic sinus tractsPre-existing searsOsteomyelitisBumsVaccination pointsImmunosuppression (organ-transplant recipients)Chemical carcinogens (arsenieals, tar)HPV5 and HPY16 infe etionTobacco useClinical featuresLesions vary from smooth nodular to verrucous, papillomatous and ulcerating lesions.All variants will eventually ulcerate as they grow.The u leers hav e a characterisecally e verted edgeDiagnosis - By biopsyManagementSurgical excisionLesion < 2 cm - Excision with 4mm marginLesion > 2 cm Excision with l cm margin

So the answer to this question is C


Instruction:
Cleavage plane sign is seen in?
A. Parosteal osteosarcoma
B. Ewing's sarcoma
C. Chondrosarcoma
D. Metastasis

Response:
Ans AParaosteal sarcoma arises exophytically from long bones and thus there is a small plane of cleavage between it and the main shaft of bone - the Cleavage plane sign.

So the answer to this question is A


Instruction:
Which of the following condition is associated with pseudopolyp formation?
A. Crohn's disease
B. Ulcerative colitis
C. Celiac disease
D. Whipple's disease

Response:
Pseudopolyps are associated with Ulcerative colitis. Other features of ulcerative colitis includes presence of erythematous, fine granular mucosal surface, in severe cases mucosa appear hemorrhagic, edematous and ulcerated. Toxic colitis or megacolon is an impoant complication. Reference: Harrison's Internal Medicine, 18th Edition, Chapter 295, Pages 2487-2489.

So the answer to this question is B


Instruction:
Pyridoxine deficiency leads to altered metabolism of?
A. Phenylalanine
B. Tryptophan
C. Methionine
D. Tyrosine

Response:
ANSWER: (B) TryptophanREF: Harper 27th ed chapter 29Tryptophan is degraded to amphibolic intermediates via the kynurenine-anthranilate pathway. Tryptophan oxygenase (tryptophan pyTrolase) opens the indole ring, incorporates molecular oxygen, and forms N-formylkynurenine. An iron porphyrin metalloprotein that is inducible in liver by adrenal corticosteroids and by tryptophan, tryptophan oxygenase is feedback-inhibited by nicotinic acid derivatives, including NADPH. Hydrolytic removal of the formyl group of N-formylkynurenine, catalyzed by kynurenine formylase, produces kynurenine. Since kynureninase requires pyridoxal phosphate, excretion of xanthurenate in response to a tryptophan load is diagnostic of vitamin B6 (pyridoxine) deficiency. Hartnup disease reflects impaired intestinal and renal transport of tryptophan and other neutral amino acids. Indole derivatives of unabsorbed tryptophan formed by intestinal bacteria are excreted. The defect limits tryptophan availability? for niacin biosynthesis and accounts for the pellagra-like signs and symptoms.

So the answer to this question is B


Instruction:
Pyruvate can be converted directly into all the following except
A. Phosphoenol pyruvate
B. Alanine
C. Acetyl CoA
D. Lactate

Response:
Phosphoenolpyruvate

While phosphoenol pyruvate can be directly converted to pyruvate (Glycolysis).
Pyruvate cannot be directly converted to phosphoenolpyruvate (Gluconeogenesis).
Conversion of phosphoenolpyruvate to pyruvate is an irreversible step in glycolysis, and is one of the barriers, in simple reversal of glycolysis.
To form phosphoenolpyruvate from pyruvate (Gluconeogenesis) pyruvate has to be first converted to oxaloacetate which is then converted to phosphoenolpyruvate by enzyme phosphoenolpyruvate carboxykinase.
Pyruvate can directly be converted to - Oxaloacetate, acetyl CoA, lactate, malate, alanine, and in yeasts and some bacteria to ethanol (by alcohol fermentation).

So the answer to this question is A


Instruction:
Pediatric airway differs from adult by
A. Large tongue
B. Sho epiglottis
C. Narrowest pa is glottis
D. Larynx in lower position

Response:
Airway: Pediatric Obligate nasal breatherLarge tongueLarynx & trachea are funnel-shapedThe narrowest pa is cricoid (glottis in adults)(Refer: Nelson's Textbook of Pediatrics, SAE, 1st edition, pg no. 2036 - 2037)

So the answer to this question is A


Instruction:
Arrange the above milestones in ascending order of appearance (starting from earliest to later milestones):
A. D - A - C - B
B. D - C - A - B
C. A - C - D - B
D. A - D - C - B

Response:
a. D - A - C - B(Ref: Nelson's 20/e p 65-77, Ghai 8/e p 45-49))A. shows mouthing that comes at 5 months.B. shows spontaneous scribbling, that appears at 15 months.C. shows Pivotting that appears around 7 months.D. shows rounded back on made to sit, that is seen at 3 months.So, the correct sequence of the given milestones is D - A - C - B.

So the answer to this question is A


Instruction:
Safe yield of a water supply source is the yield that is adequate for?
A. 80% of the year
B. 90% of the year
C. 95% of the year
D. The whole of the year

Response:
Water for meeting the requirements of a community can be obtained from various sources like rain, surface water and groundwater. The safe yield of a source of water is defined as the yield that is adequate for 95% of the year. Ref: Park's Textbook of Preventive and Social Medicine, 20th edition, p617

So the answer to this question is C


Instruction:
A 60-year-old woman has been hospitalized for three weeks with widely metastatic ovarian adenocarcinoma, and she becomes septic with vancomycin-resistant enterococcus. What is the mechanism of vancomycin resistance in this organism?
A. Acetylation of antibiotic
B. Altered drug-binding protein
C. Beta-lactamase production
D. Formation of novel cell wall peptide bridges

Response:
The antibiotic propey of vancomycin depends upon its ability to bind D-ala-D-ala, which is vital in the synthesis of peptidoglycan peptide bridges. Vancomycin-resistant enterococci instead utilize D-lactic acid in their peptide bonds, and thus are resistant to the action of vancomycin. Enzymatic deactivation of antibiotics is a common mode of resistance to aminoglycosides, chloramphenicol, and sulfonamides. The most common chemical modifications are acetylation, adenylation, and phosphorylation. Penicillin that has penetrated the cell membrane is kept in place by binding to penicillin binding proteins (PBPs). Penicillin resistance may be conferred through altered PBP affinity or number. Vancomycin resistance does not depend on PBPs. beta-lactamase is a bacterial enzyme that inactivates some of the penicillins and cephalosporins, but has no activity on vancomycin. Ref: MacDougall C. (2011). Chapter 55. Protein Synthesis Inhibitors and Miscellaneous Antibacterial Agents. 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 D


Instruction:
Regarding Lichen Planus all are true, except :
A. Hypopigmentation in most residual disease
B. Lymphatic infiltration in supradermal layer
C. Itchy polygonal, purple papule
D. Skin, hair and oral mucosa commonly involved

Response:
A. i.e. Hypopigmentation in residual disease

So the answer to this question is A


Instruction:
Gelastic seizures are seen in
A. Hypothalamic hamartoma
B. Subarachnoid tumor
C. Sheehan syndrome
D. All of the above

Response:
Gelastic seizures is burst of energy usually in form of laughing or crying occurs in hypothalamic hamartoma.

So the answer to this question is A


Instruction:
The term TUG BACK is used to indicate
A. Good fit of accessory gutta-percha point at the apical third
B. Good fit of master gutta-percha point at the apical third
C. Good fit of accessory cone in the main body of the canal
D. Good fit of the initial file at the apical third

Response:
None

So the answer to this question is B


Instruction:
Chronic dystentery, abdominal pain, and rectal prolapse in children is caused by -
A. Enterobius vermicularis
B. Ascariasis
C. Trichuris trichura
D. Trichinella spiralis

Response:
Infection with T.trichiura (trichuriasis. whipworm infection or trichocephaliasis) is asymptomatic except when the worm load is heavy. Disease may result either due to mechanical effects or allergic reaction. The worms lie threaded into the caecal mucosa and even though it is not a blood feeder, oozing of blood may occur at the sites of attachment. The blood loss is about 0.005 ml per worm per day. Over a period of time this may lead to anaemia and malnutrition. It has been suggested that mechanical blockage of the appendiceal lumen by masses of whipworms may cause acute appendicitis. In heavy infection, the worm may be abundant on the colonic mucosa, even upto the rectum. Mucus diarrhoea, chronic dysentery and abdominal pain are frequently seen in such cases. Some patients, paicularly young children may develop rectal prolapse. REF:TEXTBOOK OF MEDICAL PARASITOLOGY;CKJ PANIKER;6TH EDITION;PAGE NO165

So the answer to this question is C


Instruction:
Common presenting manifestations of Meniere's disease are all except:
A. Tinnitus
B. Vertigo
C. Sensorineural deafness
D. Loss of consciousness

Response:
(d) Loss of consciousness(Ref. Scott Brown, 8th ed., Vol 2, 819)The patient is fully conscious and oriented during an attack of Meniere.Meniere being a disturbance of peripheral vestibular system is therefore not characterised by any focal (such as diplopia, dysarthria, paresthesia, etc.) or generalised neurological symptoms like loss of consciousness, seizures, etc.The neurological symptoms accompany a central vestibular disturbance.However, vertigo of Meniere may be associated with vagal symptoms i.e. diaphoresis, pallor, nausea, vomiting, abdominal cramps, diarrhoea and bradycardia.

So the answer to this question is D


Instruction:
A 42-year-old woman is admitted to the hospital with injury to the upper (superior) trunk of the brachial plexus. The diagnosis is Erb-Duchenne palsy. Which of the following conditions is expected to be present during physical examination?
A. Winged scapula
B. Inability to laterally rotate the arm
C. Paralysis of intrinsic muscles of the hand
D. Paresthesia in the medial aspect of the arm

Response:
Injury to the superior trunk of the brachial plexus can damage nerve fibers going to the suprascapular, axillary, and musculocutaneous nerves. Damage to the suprascapular and axillary nerves causes impaired abduction and lateral rotation of the arm. Damage to the musculocutaneous nerve causes impaired flexion of the forearm. A winged scapula would be caused by damage to the long thoracic nerve. The long thoracic nerve is formed from spinal cord levels C5, C6, and C7, so the serratus anterior muscle would be weakened from the damage to C5 and C6, but the muscle would not be completely paralyzed. The intrinsic muscles of the hand are innervated by the ulnar nerve, which would most likely remain intact. Paraesthesia in the medial aspect of the arm would be caused by damage to the medial brachial cutaneous nerve (C8-T1; inferior trunk). Loss of sensation on the dorsum of the hand would be caused by damage to either the ulnar or radial nerves (C6 to T1).

So the answer to this question is B


Instruction:
Risk factors for stress urinary incontinence are all except
A. Smoking
B. Obesity
C. Hypertension
D. Pregnancy

Response:
Hypertension is not a risk factor for stress urinary incontinence.

So the answer to this question is C