Instruction:
None of the following are true about N95 masks except
A. They were designed for influenza viruses during the H1N1 outbreak
B. N- stands for National Institute of Occupational safety and Health
C. It filters 95% of all paritcles smaller than 3nm
D. A beard doesn't allow for a proper seal and makes the N95 ineffective

Response:
N 95 masks N 95 masks were designed by NIOSH for occupational safety of industrial workers. The masks come in different series N for Not resistant to oil. Used when oil paiculates are not present. R for Resistant to oil. Used when oil paiculates are present and the filter is disposed of after one shift. P for oil Proof. Used when oil paiculates are present and the filter is re-used for more than one shift. All 3 series come in variations of 95,99,100 which basically implies their percentage of filtration efficiency againt paicles bigger than 0.3 micron being 95%, 99% and 99.97% respectively. Facial hair can prevent the mask from forming a tight seal thus CDC has listed facial hairstyles which do not hamper the effectiveness of N95 These masks have been found to be effective against influenza viruses and are now currently being used for COVID-19

So the answer to this question is D


Instruction:
Most common cause of Cushing syndrome is
A. Pituitary tumor
B. Ectopic ACTH producing tumor
C. Steroid supplementation
D. Adrenal tumor

Response:
Steroid supplementation is the most common cause of question syndrome.

So the answer to this question is C


Instruction:
All are seen in congestive cardiac failure except –
A. Kerley B lines
B. Prominent lower lobe vessel
C. Pleural effusions
D. Cardiomegaly

Response:
There are prominent upper lobe vessels (not lower lobe) due to PVH.
CHF (LVF) presents radiologically with cardiomegaly and signs of pulmonary venous hypertension (see above explanation for signs of PVH).

So the answer to this question is B


Instruction:
Snyder's test tell us the:
A. Exact nature of microorganisms
B. Combined acidogenic nature of microorganisms
C. Enamel dissolving capacity of saliva
D. All of the above

Response:
None

So the answer to this question is B


Instruction:
Most common location of extradural hematoma is
A. Frontal area
B. Temporal area
C. Mastoid area
D. Posterior fossa

Response:
The most common site is temporal, as the pterion is not only the thinnest part of the skull but also overlies the largest meningeal artery –the middle meningeal. An EDH may also occur in other regions such as frontal as well as in the posterior fossa.

So the answer to this question is B


Instruction:
Cardiac output is a measure of:
A. Peripheral resistance X tissue fluid
B. Peripheral resistance X cardiac rate
C. Blood pressure X cardiac rate
D. Heart rate X blood volume

Response:
None

So the answer to this question is D


Instruction:
Intercourse in closely related individual in relation -
A. Incest
B. Adultery
C. Bestiality
D. Tribadism

Response:
Ref:Textbook of forensic medicine and toxicology (V.V.Pillay) 17th edition, page no. 404 Incest is defined as natural sexual intercourse between two persons so closely related that they are forbidden by law to marry. Incest most frequently occurs between members of nuclear family, and the victim s are most often children.

So the answer to this question is A


Instruction:
Withdrawal of heroin produces all the following symptoms except
A. Insomnia
B. Piloerection
C. Mydriasis
D. Miosis

Response:
Opioids cause miosis, withdrawal-opp. effects so not miosis Other withdrawal symptoms: yawning, insomnia, mydriasis, lacrimation, nausea, vomiting etc. opiod withdrawl is a distressing experience for the patient that makes the patient to take the sunstance again the main clue for opiod withdrawl isALL ORIFICES BLEED there is lacrimation, sweating, vomiting, rhinorrhea, diarrhoea, muscle cramps and stomach ache the piloerection that occurs in opiod withdrawl is called as COLD TURKEY the tratment of opiod withdrawl can be doneby several means SUPPOIVE CARE: clonidine+benzodiazepines+dicyclomine+anti emetic+anti diarrhoeal AGONIST ASSOSIATED DETOXIFICATION: methadone and bupenorphine ANTAGONIST ASSOSIATED DETOXIFICATION: naltrexone Ref: Essentials of postgraduate psychiatry By JN Vyas 1st ed Pg 338-340

So the answer to this question is D


Instruction:
Most potent stimulator of NaiveT-cells -
A. Mature dentritic cells
B. Follicular dentritic cells
C. Macrophages
D. B-cell

Response:
Question repeated

So the answer to this question is A


Instruction:
Least common quadrant in cancer breast is
A. Upper outer
B. Lower outer
C. Sub areolar
D. Lower inner

Response:
Most malignant type of CA breast : Inflammatory breast cancer Best prognosis is seen in : Tubular MC site of CA breast : Upper outer quadrant (left breast >right) Least common site of CA breast lower inner quadrant MC site of metastasis is Bone (osteolytic deposits in lumbar veebrae >Femur >Thoracic veebrae >Rib > Skull Ref: Sabiston 20th edition Pgno :837

So the answer to this question is D


Instruction:
All of the following are true about Brugia malayi except-
A. The intermediate hosts in India are Mansoni
B. The tail tip is free from nuclei
C. Nuclei are blurred, so counting is difficult
D. Adult worm is found in the lymphatic system

Response:
Ans. is 'b' i.e., Tail tip is from nuclei Microfilariae of B malayi . Smaller than W. bancrofti . The nuclei are blurred, hence counting is difficult. . Tail-tip is not free from nuclei. There are two discrete nuclei. . The cephalic space is longer. . The intermediate hosts for B. malayi are various species of mansonia.

So the answer to this question is B


Instruction:
Vidian neurectomy done in:
A. Allergic rhinitis
B. Vasomotor rhinitis
C. Atrophic rhinitis
D. Drug-induced rhinitis

Response:
ANSWER: (B) Vasomotor rhinitisREF: P L Dhingra 4th edition page 160, 1810%2806%2900013-3/abstractVidian neurectomy was first described for the treatment of intractable vasomotor rhinitisin the 1960s but fell into disrepute because of the recurrence of symptoms and significant complications. With subsequent improvements in the understanding of the nasal and sinus anatomy, endoscopic vidian neurectomy can now? be performed with minimal dissection and morbidityThe principle underlying vidian neurectomy is to sever the preganglionic fibers that reach the sphenopalatine ganglion through the vidian nerve. Several routes have been described to approach the deep seated vidian nerve:Trans antral Trans eptalTranspalatalTransethmoidal andTrans nasalThe most popular approach to the vidian nerve is the Transantral route.

So the answer to this question is B


Instruction:
Pathognomic of measls ?
A. Koplik spot
B. Rash
C. Fever
D. Conjuctivitis

Response:
Ans. is 'a' i.e., Koplik spot Measle Caused by RNA virus. Highly contagious droplet spread from secretion of nose and throat 4 day before and 5 days after rash. Secondary attack rate >90% in contact. Prodromal phase - characterized by fever, rhinorrhea, conjuctival congestion and dry hackig cough. Koplik spots bluish-gray specks or "grains of sand" on a red base-develop on the buccal mucosa opposite the second molars Generally appear 1-2 days before the rash and last 3-5 days Pathognomonic for measles, but not always present Rash appears on D4 first behind pinna on neck the spread of face, thrunk and abdomen. SSPE is long term complication seen in measls.

So the answer to this question is A


Instruction:
Abnormalities of bone metabolism is associated with excess of which vitamins -a) Vitamin Ab) Thiaminec) Vitamin Dd) Tocoferol
A. ad
B. c
C. ac
D. ab

Response:
None

So the answer to this question is A


Instruction:
Allosteric inhibition of an enzyme is -
A. Binding of inhibitor to catalytic site and inhibition of enzyme
B. Binding of inhibitor to other site and inhibition of enzyme
C. Inhibition of enzyme by inhibitors without binding to enzyme
D. Inactivation by phosphorylation or dephosphorylation

Response:
Ans. is 'b' i.e., Binding of inhibitor to other site and inhibition of enzyme o Some enzymes, called allosteric enzymes, posses a site, in addition to substrate binding (catalytic) site, known as the allosteric site.o Binding of allosteric modulator at the allosteric site affects the conformation of catalytic site,o Such enzymes are called allosteric enzymes.o The allosteric modulator (regulator) may facilitate the conformational change of catalytic site, required for substrate binding.o Such regulators are called allosteric activators (positive allosteric modifier); for example, fructose-2,6-bisphosphate is an allosteric activator of Phosphoffuctokinase-I.o Some allosteric regulators prevent the conformational change required for binding of the substrate,o Such regulators are called allosteric inhibitors (negative allosteric modifier); for example, citrate is an allosteric inhibitor of phosphoructokinase-I.

So the answer to this question is B


Instruction:
Commonest presentation of sickle cell anemia is -
A. Priapism
B. Bone pain
C. Fever
D. Splenomegaly

Response:
None

So the answer to this question is C


Instruction:
A newborn has a fever of 103degF. Blood culture grows gram-positive cocci in chains. This is most likely to be which of the following?
A. Group A Streptococci (Streptococcus pyogenes)
B. Group B Streptococci (Streptococcus agalactiae)
C. Salmonella species
D. Streptococcus pneumoniae

Response:
Most human infections caused by Streptococci involve the group A organisms (Streptococcus pyogenes). The group B Streptococci are members of the female genital tract and are impoant causes of neonatal sepsis and meningitis. Here we have newborn in the question given, so the most likely organism would be Gr B strep.

So the answer to this question is B


Instruction:
Herpetic keratitis is treated by
A. Analgesics
B. Atropine
C. Steroids
D. Idoxuridine

Response:
D i.e. Idoxuridine Drug of choice for herpetic keratitis is Acyclovir Q (topical) Other antiviral drugs used are - Idoxuridine - Trifluorothyrnidine - Vidarabine

So the answer to this question is D


Instruction:
True regarding estimation of Prothrombin time is:
A. Immediately refrigerate to preserve factor bility
B. Platelet rich plasma is essential
C. Run the test within 2 hours
D. Activated with Kaolin

Response:
. Run the test within 2 hours

So the answer to this question is C


Instruction:
The maximum effect of a drug is defined by-
A. Therapeutic index
B. Potency
C. Efficacy
D. Adversity

Response:
Ans. is 'c' i.e., Efficacy - Refers to maximal response that can be produced by a drug. Potency - Amount of drug needed to produce a given response.

So the answer to this question is C


Instruction:
The most common neurologic abnormality that occurs with head injury is-
A. Hemiplegia
B. Ocular nerve palsy
C. Altered consciousness
D. Convulsion

Response:
None

So the answer to this question is C


Instruction:
Which of the following is false about monoclonal gammopathy of unknown significance (MGUS)?
A. Less than 3g/dL of monoclonal protein
B. No evidence of Bence-Jones proteinuria
C. Bone marrow plasma cells less than 10%
D. Does not progress to multiple myeloma

Response:
MGUS: MGUS is the most common plasma cell disorder Asymptomatic BM Plasma Cells < 10% S. Para Protiens < 3g/dl 3% of MGUS ( aged > 50yr) can develop Multiple myeloma. 5% Of MGUS (aged >70yr) can develop multiple myeloma. Approximately 1% of patients with MGUS develop a symptomatic plasma cell neoplasm, usually multiple myeloma, per year.

So the answer to this question is D


Instruction:
Type of human papilloma virus associated with cacinoma cervix
A. Types 6, 12, 18
B. Types 16, 18, 31
C. Types 6, 8, 11
D. Types 3, 10, 19

Response:
Explanation:
Juvenile Papillomatosis of the breast
● It is a relatively common benign localised proliferative lesion in the breast.
● It is mainly seen in young women (mean age: 19-23 years)
● Caused by HPV 6 and 11.
Clinical presentation
● Patients present with a firmt well-defined, mobile mass often in the periphery of the breast.
● There is usually no nipple discharge.
Pathology
● Papillary proliferation of the ductal epithelium which partly fills up smaller ducts causing distension.
● Grossly, well-circumscribed mass containing multiple small cysts (<2 cm) within a dense fibrous stroma
(therefore sometimes termed swiss cheese disease by pathologists).
● Lesions can vary in size, usually ranging from 1 to 8 cm.
Radiographic features
Breast ultrasound
● They can appear as an ill-defined, inhomogenous hypoechoic mass with multiple small (up to 4mm) predominantly peripheral cysts.
● Microcalcifications may be seen at sonography.
Mammography
● These lesions are usually negative on mammography.
● Occasionally may show microcalcifications, an asymmetric density or a prominant intraductal pattern.
Galactography
● May show a multiple irregular filling defects within the breasts. Breast MRI
Signal characteristics according to one report include
● T1 : hypo-intense lobulated mass
● T2 : may show- presence of multiple small internal cysts, best shown on this sequence and is considered the most specific feature
● T1 C+ (Gd) : may or may not show marked rapid enhancement
● Dynamic sequence: shows a benign enhancement profile.
Complications
● Despite being a benign entity, it is considered by some to be a marker for familial breast cancer.
● Approximately /0% develop breast cancer later in life.

So the answer to this question is B


Instruction:
The tertiary structure of protein is detected by
A. X-ray diffraction/crystallographs.
B. Spectrophotometry
C. Electrophoresis
D. Chromatography

Response:
None

So the answer to this question is A


Instruction:
True about ALL –
A. Anemia
B. Leukopenia
C. Hepatomegaly
D. All of the above

Response:
Clinical manifestations of ALL

Symptoms related to depression of normal marrow function.
An accumulation of neoplastic blast cells in the bone marrow suppresses normal hematopoiesis by physical crowding and competition for growth factors.
This results in : -

Anemia → Fatigue, pallor
Neutropenia→Infection, fever
Throtnbocytopenia → Bleeding, petechiae, ecchymoses, epistaxis.

These are the most common sign & symptoms.

Bone pain and tenderness

Resulting from marrow expansion and infiltration of the subperiosteum.

Generalized lymphadenopathy, splenomegaly and hepatomegaly

Caused by neoplastic infiltration of these tissues.
More common in ALL than AML.

Symptoms related to compression of large mediastinal vessels or airway

In Pre-T ALL of thymus.

CNS manifestation

Due to meningeal involvement.
A headache, vomiting, nerve palsies.
More common in ALL than in AML.

Testicular involvement is also more common in ALL than in AML

So the answer to this question is D


Instruction:
All are principles of primary health care except?
A. Intersectoral coordination
B. Community paicipation
C. Appropriate technology
D. Decentralised approach

Response:
o There are 4 main principles of primary health care : 1) Equitable distribution 2)Community paicipation 3)Intersectorial coordination 4) A ppro priate technology Ref: Park 24th edition

So the answer to this question is D


Instruction:
Which of the following agent is not used in erectile dysfunction?
A. PGE2
B. Vardenafil
C. Phenylephrine
D. Alprostadil

Response:
Ans. c. Phenylephrine (Ref: KDT 7/e p303-304, 6/e p295-296; Harrison 19/e p326, 18/e p377)PGE2, Vardenafil (phosphodiesterase-V inhibitor) and Alprostadil (PGE1) are used in erectile dysfunction.Phenylephrine is not used in erectile dysfunction.L-arginine - increased NO - Increases c-GMP - Decreased Ca2+ - Smooth muscle relaxation - ErectionTreatment of Erectile DysfunctionSildenafilOnly approved and effective oral agents for erectile dysfunctionQUseful in wide range like in psychogenic, diabetogenic, vasculogenic, post radical prostatectomy (nerve sparing procedure) and spinal cord injury.Acts by inhibiting phosphodiesterase V (PDE -V)Q and releasing NOQ from nerve endings and endothelium.C-GMP is metabolized by PDE-V which is inhibited by sildenafil.Other phosphodiesterase-V inhibitors: TadalafliQ and vardenafilQPIPE therapy: (PapaverineQ/Phentolamme induced penile erection)AndrogenQIntra-urethral alprostodil (PGE1)QPenile prosthesis, vacuum constriction deviceRemember:Master Johnson technique is for premature ejaculationQ.Beta blockers can cause erectile dysfunctionQ.Drugs causing Erectile DysfunctionEnzyme Inhibitors: Ketoconazole, cimetidineAll Anti-hypertensives: Minimum with ACE inhibitors, maximum with beta-blockersAll abusive drugs at high doseFemale hormonal drug given to male, e.g., SERM, estrogen, exmestaneAll psychotic drugsAnti-depressants and anti-psychotic drugs

So the answer to this question is C


Instruction:
Nuclear cytoplasmic ratio in malignant cells is
A. 1:06
B. 1:05
C. 1:01
D. 6:04

Response:
None

So the answer to this question is C


Instruction:
All of the following lesions may be classified as
odontogenic tumours EXCEPT
A. Acanthomatous ameloblastoma
B. Branchial cleft cyst
C. Myxoma
D. Simple ameloblastoma

Response:
None

So the answer to this question is B


Instruction:
Which of the following is not a cause for interstitial keratitis
A. Syphillis
B. Leprosy
C. Tuberculosis
D. Pneumococcus

Response:
Pneumococcus Interstitial keratitis Corneal stromal inflammation without epithelial involvement. Infective causes are: *Syphillis * TB * Leprosy *Acanthamoeba

So the answer to this question is D


Instruction:
In human corneal transplantation, the donor tissue is
A. Synthetic polymer
B. Donated human cadaver eyes
C. Donated eyes from live human beings
D. Monkey eyes

Response:
B i.e. Donated human cadaver eyes Donor corneas are harvested from cadaveric donors within 6 hours after death Q (upto 12 hours after death in cold countries.) Corneal Transplantation (Grafting) /Keratoplasty General Indications To provide a clear visual axis and restore good vision (Optical Keratoplasty) To restore the integrity of the globe in corneal diseases (Tectonic Graft) To remove infective tissue unresponsive to conservative treatment (Therapeutic Graft) To improve the appearance of the eye (Cosmetic Graft) Contraindications Death of unknown cause CNS disease such as Creutzfeldt-Jacob disease, subacute sclerosing panencephalitis, rubella, Reye's syndrome, rabies & infectious encephalitis. Infections such as HIV, Hepatitis, septicemia, syphilis, & endocarditis. Eye disease as retinoblastoma, malignant tumor of anterior segment & active ocular inflammation (uveitis, scleritis, retinitis & choroiditis) Prior ocular surgery (although pseudophakic eyes may be used with good cell density) Congenital or acquired anterior segment abnormalities such as Keratoconus & Fuch's endothelial dystrophy. * Corneas from infants (53 year) are used only very occasionally, even for pediatiric transplants as they are a/w surgical, refractive and rejection problems Graft-Harvesting Donor corneas are harvested from cadaveric donors within 6 hours after death Q (upto 12 hours after death in cold countries.) Graft-Preservation Sho term (upto 96 hours) - Moist chamber (at +40C) method for 24 hours. - M-K (Mc Carey-Kaufman) medium upto 96 hours. Intermediate-term (upto 2 weeks) - K-SOL / DEXOL / OPTISOL-medium Long term (months - years) - Viable: Organ culture & cryopreservation - Nonble: Glycerine Types Full Thickness or Penetrating keratoplasty Paial Thickness or Lamellar Keratoplasty - Superficial LK (replacement of corneal epithelium & superficial stroma) Deep anterior LK (DALK). all opaque tissue almost upto the level of Descemet membrane is removed Descemet stripping endothelial Keratoplasty (DSEK)/DS automated EK (DSAEK; when a microkeratome is used to perform donor lamellar dissection)/ Posterior LK (PLK) Small Patch grafts which can again be full or paial thickness Full Thickness/ Penetrating Keratoplasty (PKP) It is full thickness replacement of diseased corneal tissue with a healthy donor. It can be done in disease involving all or few layers of cornea eg keratoconus, pseudophakic bullous keratopathy, Fuchs endothelial & other dystrophies. An ideal graft size is 7.5 mm; grafts smaller than this may 1/t high astigmatism. Grafts 8.5 mm are prone to postoperative anterior synechiae formation, vascularization and increased intraocular pressure. - PKP has long been the gold standard for treatment of endothelial dysfunction. The limitations of PK include However, because of recent advances & lack of these limitations, endothelial keratoplasty is rapidly gaining popularity. It takes 6 months to many years for the refraction to stabilize 10-15% of patients typically require a hard contact lens for best vision and final mean refractive cylinder of 45D is common PKP incision cuts all corneal nerves, so the inclination to blink and produce tears is reduced postoperatively. This, together with prolonged presence of corneal sutures, increases the risk that ocular surface complications will interfere with recovery. PKP wound never heals back to the full strength of a virgin cornea, so a PKP eye is forever at increased risk of loss from a traumatic injury. Paial Thickness /Lamellar Keratoplasty (LKP) - It is a procedure in which a paial thickess graft of donor tissue is used to provide tectonic stability and /or optical improvement. - It may be of 2 types: anterior LKP and posterior LKP. - In anterior LKP, the transplanted tissue does not include corneal endothelium. This procedure avoids endothelial rejection and thus donor tissue may be obtained from older eyes. Indications for anterior LKP mainly include anterior corneal pathology in which the posterior cornea is unaffected - In deep LKP and posterior LKP, the diseased corneal endothelium is replaced while keeping the anterior corneal surface intact. - Because the donor endothelium is not used, the criteria for anterior LKP are less stringent than those used in PKP; the tissue does not need to be as fresh as that used in PKP. Corneal stroma may be used upto 7 days postmoem (whereas, Posterior LKP has same criteria as PKP) . Endothelial Keratoplasty (EK) i.e. posterior lamellar keratoplasty/ Descemet's Stripping endothelial Keratoplasty involves selective removal of dysfunctional recipient corneal endothelium and replacement with donor tissue consisting of posterior stroma and healthy endothelium. EK is performed through a small incision and spares the majority of host cornea so corneal strength and surface topography are minimally altered & technique is essentially refractive neutral. Fuhermore, corneal innervaton is retained and corneal sutures are not required. So corneal surface complications are minimal. The small incison allows rapid healing and rapid visual recovery after surgery. While fewer patients may achieve 20/20 vision after EK compared with PKP, overall the visual results are more predectible. - DSEK increases corneal thickness because posterior donor stroma is implanted without removal of any recipient stoma. In general, younger patients are likely to achieve better visual acuity after DSEK than older patients (even when have no retinal problems). Whereas PKP was often post phoned until after retirement or at least until visual problems were quite disturbing, EK is now being performed earlier, when visual problems begin to interfere with daily activities, such as reading or driving Infact, it is preferable to perform EK before long standing corneal edema 1/ anterior stromal scarring, because the anterior stroma is not replaced in EK.

So the answer to this question is B


Instruction:
Most common form of glomerulonephritis worldwide is?
A. IgA nephropathy
B. Post streptococcal glomerulonephritis
C. Rapidly progressive glomerulonephritis
D. Acute glomerulonephritis

Response:
Most Common GN - IgA nephropathy / Berger disease (MC C/F - gross Microscope Hematuria) Most Common Nephrotic overall - FSGS Most Common Nephrotic in adult - FSGS Most Common Nephrotic in Children - Minimal Changes disease Most Common Nephrotic in elderly - MGN

So the answer to this question is A


Instruction:
The following vitamin acts as a cofactor for glycogen phosphorylase:
A. Thiamine pyrophosphate (B1)
B. Riboflavin (B2)
C. Pyridoxal phosphate (B6)
D. Methyl cobalamin (B12)

Response:
Rate limiting enzyme of glycogenolysis is glycogen Phosphorylase which requires PLP i.e. vitamin B6. This vitamin is required as donor of phosphate as Glycogen phosphorylase transfers phosphate to glucose, releasing glucose-1-phosphate. Pyridoxal phosphate is also cofactor for transamination and decarboxylation reaction. Here is a difference to be kept in mind. Transamination Decarboxylation Aldehyde group of pyridoxal phosphate (PLP) plays a role Phosphate group of PLP plays a role

So the answer to this question is C


Instruction:
Which of the following agents do not act GABAA Cl- channel complex receptors?
A. Zopiclone
B. Benzodiazepines
C. Thiopentone
D. Promethazine

Response:
Zopiclone (non benzodiazepine hypnotics ),benzodiazepine ,thiopentone(barbiturates) all act on GABA-Cl channel ,whereas promethazin is a antihistamine drug that acts on histaminergic receptors which causes high sedation. Ref: KD Tripathio 8th ed.

So the answer to this question is D


Instruction:
Which complement component is anaphylotoxin -
A. C3b
B. C5.-9
C. C5a
D. Cl-3

Response:
Ans. is 'c' i.e., C5a

So the answer to this question is C


Instruction:
Which of the following screening methods is used under RNTCP?
A. Active
B. Passive
C. Mass
D. All of the above

Response:
Under RNTCP, active case finding is not pursued. Case finding is passive. Patients presenting themselves with symptoms suggesting tuberculosis are screened with two sputum smear examinations. Ref: Park 21st edition, page 381.

So the answer to this question is B


Instruction:
Down syndrome most commonly occurs due to ____
A. Reciprocal translocation
B. Non-disjunction in maternal meiosis
C. Mosaicism
D. Non-disjunction in paternal meiosis

Response:
Down&;s syndrome occurs due to nondisjunction in maternal meiosis. Down&;s syndrome: Most common chromosomal disorder Incidence: 1/800 live bihs Mechanism: Trisomy 21 (meiotic maternal nondisjunction) (95%) Robesonian translocation: 4% Mosaic pattern: 1% Non-disjunction points to the failure of the homologous chromosome to seperate during the meiotic division. Ref: Nelson textbook of pediatrics 21st edition Pgno: 658

So the answer to this question is B


Instruction:
Masculine voice in females is treated by
A. Thyroplasty type 1
B. Thyroplasty type 2
C. Thyroplasty type 3
D. Thyroplasty type 4

Response:
The masculine voice in females is treated by Thyroplasty type 3. It is used to shoen (relax) the vocal cord. Relaxation of vocal cord lowers the pitch. This procedure is done in mutational falsetto or in those who have undergone gender transformation from female to male. Ref: PL Dhingra Textbook of Ear, Nose and Throat, Edition 6, page - 302

So the answer to this question is C


Instruction:
Cerebellar hemangioblastoma and retinal tumors are seen in:
A. Von Hippel Lindau syndrome
B. Neurofibromatosis type 1
C. Tuberous sclerosis
D. Neurofibromatosis type 2

Response:
Ref: Chapter 28. CNS. page - 1343: Robbins and Cot ran Pathologic Basis of Disease: 8th editionExplanation:The Von Hippel Lindau gene (VHL) is tumor suppressor gene located on Chromosome 3p25-p26 and down-regulates expression of vascular endothelial growth factor; hence its dysregulation is associated with hemangioblastomas and retinal tumors.The neurofibromatosis type 1 (Chr. 17) is associated with:Optic nerve gliomasLisch nodules (pigmented nodules of the iris)Cafe au lait spots (cutaneous hyperpig- mented macules)Neurofibromatosis type 2 (Chr. 22) is associated with:8th nerve schwannomasMultiple meningiomasGliomasEpendymoma of spinal cordTuberous sclerosis complex: It is associated with:Cortical tubersSubependymal nodulesSubependymal giant cell astrocytomasRenal angiomyolipomasRetinal glial hamartomasPulmonary lymphangioleiomyomatosisCardiac rhabdomyomasShagreen patches (cutaneous leathery thickenings)Ash-leaf patches (hypopigmented areas)Subungual fibromas

So the answer to this question is A


Instruction:
Double bleb sign in USg are depictive of:
A. Intrauterine two gestations sac
B. Amniotic sac and yolk sac
C. Ectopic pregnancy
D. Heterotopic pregnancy

Response:
Double bleb sign : Visualization of gestational sac containing a yolk sac and amniotic sac giving an appearance of two small bubbles . Feature of an intrauterine pregnancy. Distinguishes a pregnancy from a pseudogestational sac or decidual cast cyst. Ref:Williams 24e pg 170.

So the answer to this question is B


Instruction:
Tolvaptan is used for:-
A. SIADH
B. Central DI
C. Von Wilebrand disease
D. Catecholamine resistant Shock

Response:
In SIADH, ADH increases so more activation of V2 receptor, so more water absorption. The patients suffer from dilutional hyponatremia. Option B: Managed by Desmopressin / Nasal spray Option C: Managed by Cryoprecipitate / VW Factor Option D: No definitive treatment, patient dies more often

So the answer to this question is A


Instruction:
Which of the following drugs should not be given in renal failure ?
A. Clindamycin
B. Methicillin
C. Amoxicillin
D. Rifampicin

Response:
None

So the answer to this question is B


Instruction:
Which one of the following modes of intervention covers the process of advising a patient to change his lifestyle in order to prevent any risk factor for a disease
A. Health promotion
B. Specific protection
C. Disability limitation
D. Rehabilitation

Response:
Lifestyle modification (modification of risk factors) comes under health promotion which in turn is a pa of primary prevention Primary prevention includes both health promotion and specific protection Ref : Park&;s Textbook of Preventive and Social Medicine; 23rd edition - Pgno. 42

So the answer to this question is A


Instruction:
A-7 year child presenting with acute otitis media, does not respond to ampicillin. Examination reveals full and bulging tympanic membrane, the treatment of choice is -
A. Systemic steroid
B. Ciprofloxacin
C. Myringotomy
D. Cortical mastiodectomy

Response:
None

So the answer to this question is C


Instruction:
Which of the following compound antagonizes the actions of insulin
A. Neuropeptide Y
B. Growth hormone
C. Substance P
D. Vasoactive intestinal peptide

Response:
Ans. (b) Growth hormone(Ref: Ganong, 25th ed/p.434)Growth hormone is a diabetogenic hormone. It causes hyperglycemia. It antagonizes the action of insulin

So the answer to this question is B


Instruction:
A wound specimen obtained from a person working with wool from a Caribbean island demonstrated a large gram-positive rod from a non- hemolytic colony with swirling projections on blood agar. The most likely method to demonstrate spores would be which of the following
A. Gram stain
B. Acid-fast stain
C. India ink stain
D. Malachite green stain

Response:
Spores can be observed as intracellular refractile bodies in unstained cell suspensions. Also, they are commonly observed by staining with malachite green or carbolfuchsin. The spore wall is relatively impermeable, but heating of the preparation allows dyes to penetrate. Alcohol treatment then serves to prevent spore decolorization. Finally, the spores are counterstained.

So the answer to this question is D


Instruction:
lining mucosa is found over all of the following surface except:
A. Cheeks.
B. Floor of the mouth.
C. Ventral surface of tongue.
D. Dorsum of tongue.

Response:
Dorsum of the tongue is specialized mucosa

So the answer to this question is D


Instruction:
Which of the following is feature of Alzheimer's disease
A. Early memory loss
B. Recent memory loss
C. Recent and past memory loss
D. Global memory loss

Response:
Recent memory loss

So the answer to this question is B


Instruction:
About CSF following statement (s) is/are true
A. The dura is attached firmly to bone
B. CSF has a net weight of 1400g
C. Total volume of CSF is 150 ml
D. Daily production is 550 ml/day

Response:
CSF has a net weight of only 50 g In humans, the volume of CSF is about 150 mLThe rate of CSF production is about 550 mL/day. The dura is attached firmly to boneRef: Ganong's Review of Medical Physiology Twenty-Third Edition Page No: 572

So the answer to this question is B


Instruction:
Which is elaborated by small cell carcinoma lung?
A. PTH
B. ACTH
C. Erythropoesitn
D. Noradrenaline

Response:
Ans. is 'b' i.e., ACTH o The ectopic hormones produced by lung carcinoma are:# ACTH} # AAVP(ADH)- Produced by Small cell Ca# ANF# Calcitonin # PTH- Produced by Non-small cell Ca

So the answer to this question is B


Instruction:
Punishment for a person seeking prenatal diagnostic facilities for purpose of sex selection is
A. 1 year imprisonment and fine Rs.10000
B. 3 - 5 years imprisonment and fine Rs.50000 - 1 lakh
C. 1 year imprisonment and fine Rs.50000 - 1 lakh
D. 3 year imprisonment and fine Rs.10000

Response:
Section 23(3) PCPNDT ACT (Pre-Conception and Pre-Natal Diagnostic Techniques (PCPNDT)Act, 1994) Punishment for a person seeking prenatal diagnostic facilities for purpose of sex selection is 3 - 5 years imprisonment and fine Rs.50000 - 1 lakh

So the answer to this question is B


Instruction:
Proteins are soed by:
A. Golgi Bodies
B. Mitochondria
C. Ribosomes
D. Nuclear Membrane

Response:
The Golgi apparatus is involved in the glycosylation and soing of proteins. Ref: Biochemistry By Mary K. Campbell, Shawn O. Farrell, 7th Edition, Page 21; Harper's Textbook of Biochemistry, 27th Edition, Pages 506, 518

So the answer to this question is A


Instruction:
The virus with smallest genome is ?
A. Reovirus
B. Parvovirus
C. Picornavirus
D. HIV

Response:
Ans. is 'b' i.e., Parvovirus . Smallest genome ----> Parvovirus . Largest genome --> Poxvirus

So the answer to this question is B


Instruction:
A child presents with recurrent chest infections and recurrent sinusitis. Chest X-ray reveals dextrocardia and situs inversus. The most likely diagnosis is:
A. Kartagener's syndrome
B. Ehler-Danlos syndrome
C. Marfan's symdrome
D. William Campbell syndrome

Response:
Ans, a. Kartagener's syndrome (Ref: Robbins 9/e p683; Nelson 19/e pi 49?)The most likely diagnosis in a child with history of recurrent chest infections and recurrent sinusitis with dextrocardia and situs inversus on chest X-ray is Kartagener's syndrome.Kartagener's Syndrome (Immotile cilia syndrome/Primary ciliary dyskinesia)* Primary ciliary dyskinesia (PCD) comprises those respiratory disorders having in common the malfunction of airway cilia.* The abnormality results from various inherited primary structural defects in the cilia that lead to repeated and chronic lung and sinus infections.* About 50% of patients with PCD have Kartagener's syndrome: Situs inversus, chronic sinusitis and otitis, and airway disease leading to bronchiectasis.* Recurrent sinopulmonary infections are seenKartagener's Syndrome (Immotile cilia syndrome/Primary ciliary dyskinesia)Ultrastructural changes causing immotility of ciliaImmotility during embryogenesisImmotility in respiratory EpitheliumImmotility of cilia in sperms* Situs inversus* Dextrocardia* Bronchiectasis* Sinusitis* Infertility (male) Major cardiac AnomalyMajor non cardiac anomalyKartagener's syndrome (Immotile cilia syndrome) (Primary ciliary dyskinesia)* Dextrocardia0* Situs inversus* Sinusitis* Bronchiectasis* Infertility

So the answer to this question is A


Instruction:
Which is not visualized on posterior rhinoscopy:
A. Eustachian tube
B. Inferior meatus
C. Middle meatus
D. Superior concha

Response:
Posterior rhinoscopy: It is method of examination of the posterior aspect of nose and pharynx. Structures seen on posterior rhinoscopy: - Both choanae - Posterior end of nasal septum - Opening of Eustachian - Posterior end of superior/tube middle and inferior turbinates - Fossa of Rosenmuller - Torus Tubarius - Adenoids - Roof and posterior wall and nasopharynx.

So the answer to this question is B


Instruction:
Immune complex deposition is the basic underlying pathology in which type of hypersensitivity reaction?
A. Type I
B. Type 2
C. Type 3
D. Type 4

Response:
Ref: Robbins Pathologic Basis of Disease. 8flt editionExplanation:(See the following table)

So the answer to this question is C


Instruction:
According to Naegele's rule, calculate EDD of a patient with LMP 9/1/2017:
A. 16/10/2017
B. 16/09/2017
C. 16/11/2017
D. 9/10/2017

Response:
ANS. A* Adding 7 days + 9 months to LMP gives EDD* Naegele's rule is based on 28 days regular cycle* If the cycle is shorter or longer than 28 days, EDD will be corrected and written as corrected EDD.For irregular cycles:1. If she is having 40 days cycle regularly, to get corrected EDD, add (40-28) with the EDD calculated from LMP.2. If she is having 21 days cycle regularly, to get corrected EDD, substract (28-21) 7 with the EDD calculated from LMP.

So the answer to this question is A


Instruction:
To avoid pocket formation distance between the finish line and alveolar crest should be:
A. 0.5 mm
B. 1 mm
C. 1.5 mm
D. 2.5 mm

Response:
If the distance between finish line and the alveolar crest is less than 2mm, the restoration may lead to gingival. inflammation, loss of alveolar crest and pocket formation.

So the answer to this question is D


Instruction:
Subodontoblastic plexus of Raschkow occupies
A. Cell rich zone
B. Cell free zone
C. Odontoblastic zone
D. All of the above

Response:
None

So the answer to this question is B


Instruction:
Low risk type of HPV -
A. Type -16
B. Type - 6
C. Type-18
D. Type-31

Response:
Ans. is 'b' i.e., type 6 o High risk HPV: HPV-16,18,31,33 & 45 - associated with cervical cancero Low risk HPV: HPV- 6 & 11 - associated with precursor lesions CINo HPV-5 & 8 - associated with squamous cell cancer in patients with epidermodysplasia verruciformis.

So the answer to this question is B


Instruction:
The total number of germ tooth present in a full form foetus is
A. 20
B. 28
C. 44
D. 48

Response:
Total number of tooth germs formed in human fetus is 52 (20 primary + 32 permanent). Third molar tooth germ initiated at 4 or 5 years. If you exclude the third molar tooth germs, the answer will be 48.

So the answer to this question is D


Instruction:
Cardiomyopathy may be seen in all of the following except :
A. Duchene muscular dystrophy
B. Friedrich's ataxia
C. Type II glycogen storage disease
D. Alkaptonuria

Response:
Answer is D (Alkaptonuria) Friedreich's ataxia, Glycogen storage diseases and Muscular dystrophies are known to cause of cardiomyopathy. 11kantonuria hay not been mentioned to cause cardiomi'opathv. Primary myocardial involvement Idiopathic (D,R,H) Failmilial (D,R,H) Eosinophilic endomyocardial disease (R) Endomyocardial fibrosis (R) Secondary Myocardial involvement Infective (D) Connective tissue disorder (D) * Viral myocarditis * Systemic lupus erythematosus * Bacterial myocarditis * Polyaeritis nodosa * Fungal myocarditis * Rheumatoid ahritis * PRotozoal myocarditis * Progressive systemic sclerosis * Metazoal myocarditis * Dermatomyositis * Spirochetal Infiltratiions and granulomas (R,D) * Rickettsial * Amyloidosis Metabolic (D) * Sarcoidosis Familial storage disease (D,R) * Malignancy * Glycogen storage disease Neuromuscular (D) * Mucopolysaccharidoses * Muscular dystrophy * Hemochromatosis * Myotonic dystrophy * Fabry's disease * Friedreich's ataxia (H,D) Deficiency (D) Sensitivity and toxic reactions (D) * Electrolytes * Alcohol * Nutritional * Radiation * Drugs Peripaum hea disease (D)

So the answer to this question is D


Instruction:
A 27 yr old worker fall from a roof and is brought to the emergency depament. His physical examination and computed tomography (CT) scan reveal dislocation fracture of the thoracic veebrae. The fractured body of the T7 veebra aiculates with which of the following pas of the ribs?
A. Head of the sixth rib
B. Neck of the seventh rib
C. Tubercle of the seventh rib
D. Head of the eighth rib

Response:
The body of veebra T7 aiculates with the head of the seventh and eighth ribs. The neck of a rib does not aiculate with any pas of the veebra. The transverse process of the veebra aiculates with the tubercle of the corresponding rib.

So the answer to this question is D


Instruction:
Mechanism of action of d-tubocurarine is
A. Competitive, nondepolarizing block at the Nm cholinergic receptor
B. Non - competitive, nondepolarizing block at the Nm cholinergic receptor
C. Competitive, depolarizing block at the Nm cholinergic receptor
D. Noncompetitive, depolarizing block at the Nm cholinergic receptor

Response:
Depolarizing MR - Succinylcholine | - Shoest acting MR - C/I in nerve and muscle injury because it can cause | K+ - Hypehermia can occur. - It causes fasciculations responsible for post-operative muscle pain. Non - Depolarizing or competitive MR d- Tubocurarine - It releases lot of histamine that leads to vasodilation and bronchoconstriction. It competitively inhibit the Nm receptor.

So the answer to this question is A


Instruction:
Dental caries is associated with flouride levels in water below:
A. 0.1 mg
B. 0.5 mg /L.
C. 1.0 mg /L
D. 2.0 mg /L

Response:
None

So the answer to this question is C


Instruction:
Treatment of severe ulnar neuritis in borderline tuberculoid leprosy is which of the following?
A. MDT only
B. MDT +steroid
C. Wait and watch
D. MDT + thalidomide

Response:
Borderline tuberculoid leprosy patients may experience both type 1 ant type 2 lepra reactions. If there is nerve involvement in type 1 lepra reaction, coicosteroids are added to the Multi drug treatment regimen. Ref: Textbook of Dermatology, Venereology and Leprology, Devinder Mohan Thappa, 2nd Edition, Page 326 & 332; Handbook of leprosy by Jopling & Mc Dougall 5,th /127, 123; Current treatment options in Neurology 2002, / 189- 196.

So the answer to this question is B


Instruction:
What is the gold standard investigation of choice for deep vein thrombosis?
A. Double contrast duplex
B. Venography
C. CT scan
D. MRI

Response:
Ans. B. Venography* Duplex ultrasound is the initial investigation of choice in nearly all patients with suspected DVT.* Its reliability is dependent upon the skill of the user.* Major axial veins of the lower limb are well displayed.* It has a sensitivity of 98.7% and specificity of 100% for above-knee DVT and a sensitivity of 85.2% and specificity of 98.2% for below-knee DVT.* Gold standard/Best test for venous occlusion is invasive venography.

So the answer to this question is B


Instruction:
Which of the following is the most common cause of Meningoencephalitis in children
A. Mumps
B. Arbovirus
C. HSV
D. Enterovirus

Response:
Most common cause of meningoencephalitis in children is enterovirus. Meningoencephalitis is mainly caused by enteroviruses, varicella-zoster viruses, herpes simplex viruses, or measles viruses. Reference: GHAI Essential pediatrics, 8th edition

So the answer to this question is D


Instruction:
Embryo is termed till:
A. 12 weeks after LMP
B. 10 weeks after fertilization
C. 10 weeks after LMP
D. 8 weeks of fertilization

Response:
Ans. (d) 8 weeks of fertilizationRef. Dutta's Obstetrics 7th ed./ 41The prenatal period can be divided intoPeriodProduct (k/a)Extends* Ovular/GerminalOvum1st 2 weeks after ovulation* EmbryonicEmbryoFrom fertilization (2nd week of gestation) to 8th week of development of fertilization (10th week AOG)* FetalFetusFrom 8th week of fertilization or development (10th week AOG to delivery)

So the answer to this question is D


Instruction:
Skin over hypothenar eminence is supplied by?
A. Radial nerve
B. Median nerve
C. Anterior interosseous nerve
D. Ulnar nerve

Response:
Ulnar nerve

So the answer to this question is D


Instruction:
Projections from substantia nigra pars compacta to striatum releases
A. GABA
B. Dopamine
C. Glutamate
D. Aspartate

Response:
None

So the answer to this question is B


Instruction:
Rhinitis most common bacterial cause -
A. Haemophilus influenza
B. Streptococcus haemolyticus
C. Pasturellamultocida
D. Comybacterium diphtheria

Response:
Ans. is 'a' i.e., Haemophilus influenza o Acute bacterial rhinitis is most commonly seen among children, but adult may develop the condition after nasal trauma, viral upper respiratory tract infection, or surgery,o The clinical presentation of acute bacterial rhinitis may be identical to that of common cold,o Most common causative organisms include S. pneumoniae, H. influenzae and Moraxella Catarrhalis.o Note: Overall, most common cause of infective rhinitis is viral infection (viral rhinitis).

So the answer to this question is A


Instruction:
PDL is thinnest at:
A. CEJ
B. Apex
C. Middle
D. None of the above

Response:
None

So the answer to this question is C


Instruction:
A child presents with hypochromic microcytic anemia, with normal levels of free Erythrocyte Protoporphyrin The most likely diagnosis is:
A. Iron deficiency anemia
B. Lead toxicity
C. Thalassemia
D. Anemia of chronic disease

Response:
Answer is C (Thalassemia) Erythrocyte protoporphyrin levels are characteristically normal in Thalassemia Free Erythrocyte Protoporphyrin is a useful screening test for microcytic hypochromic anemias. Free Erythrocyte protoporphyrin levels are increased in iron deficiency anemias, lead poisoning many cases of sideroblastic anemia and anemias of chronic disease. Erythrocyte protoporphyrin levels are normal in Thalassemias. Parameter Iron Deficiency Lead Poisoning Thalassemia trait Chronic disease Free Erythrocyte Ted Ted Normal Ted Protoporphyrin (FEP)

So the answer to this question is C


Instruction:
Anaesthetic that has a smooth induction is –
A. Diethyl ether
B. Thiopental
C. N2O
D. Halothane

Response:
This question is straight forward as amongst the given options only thiopental is used as inducing agent.
Iv. inducing agents are used for induction because of rapid and smooth induction and drugs used for this purpose are:

Thiopentone
Methohexitone sod
Propofol
Etomidate

Till now we were discussing about inducing property of iv inducing agents. Some of the inhalation anaesthetic agents also have smooth induction.
Induction by inhalation agents
Unpleasant

Ether

Intermediate

Halothane
Enflurane

Smooth

Isoflurane
Desflurane
Sevoflurane
Nitrous oxide

But, you should keep in mind that smoothest induction is produced by iv inducing agents and therefore they are used as inducing agent.

So the answer to this question is B


Instruction:
A 64-year-old man developed persistent back pain. On examination, his BP is 120/80, pulse 70, respiratory rate is 15, and temp is 98.6degF. Physical findings include severe pallor & evidence of muscle wasting. On urinalysis, 4 + protein. M/E of urine reveals occasional broad and occasional granular cast. Lab studies: Serum creatinine = 5.1 mg/dl, serum Na = 141, K = 5.6, chloride = 101, CO2 = 14, serum calcium = 11.7, and serum phosphorus = 6.0. Most likely etiology is:
A. Renovascular disease
B. Thrombotic renal disease
C. . Multiple myeloma
D. SLE

Response:
Multiple myeloma Multiple myeloma is the most likely etiology. The combination of hypercalcemia and acute renal failure raises the possibility of multiple myeloma as the bone breakdown secondary to tumor involvement releases large amounts of calcium to the extracellular fluid and hypercalcemia ensues. The renal failure in myeloma is primarily related to hypercalcemia combined with proteinaceous cast formation within the renal tubules, producing a form of intratubu- lar obstruction as well as a tubular inflammatory lesion. The major diagnostic clue is the finding of dipstick mildly positive urine with a sulfosalicylic acid strongly positive urine. Dipstick testing does not detect the negatively charged light-chain proteins, only the alburnin. Renovascular lesions and thrombotic renal disease could present with this picture, although they should not be asso- ciated with hypercalcemia and severe back pain, and findings upon examination of the urine would not include proteins. Systemic lupus erythematosus (SLE) can, of course, be associated with severe anemia and joint mani- festations, but hypercalcemia is not pa of the picture.

So the answer to this question is C


Instruction:
Which is most specific for Alcholic liver disease
A. AST
B. ALT
C. GGT
D. ALP

Response:
AST: ALT ratio of  3:1 is highly specific
Next is GGT

So the answer to this question is C


Instruction:
Tetracycline chelates with
A. Ca salts
B. Mg salts
C. Al salts
D. All of the above

Response:
None

So the answer to this question is D


Instruction:
Which of following pregnancy complications is associated with polycythemia vera
A. Coagulopathy
B. Placental abruption
C. Stillbirth
D. Placenta previa

Response:
Polycythemia vera → Increased blood viscosity → Thrombotic complications and fetal loss.

So the answer to this question is C


Instruction:
Most potent stimulus for secretin is -
A. Dilatation of intestine
B. Acid chime
C. Protein
D. Fat

Response:
B i.e. Acid chyme

So the answer to this question is B


Instruction:
Acute flaccid complete paralysis with areflexia and loss of perianal reflexes, below the level of spinal cord injury is due to -
A. Spinal shock
B. Denervation
C. Malingering
D. UMN paralysis

Response:
Ans. is 'a' i.e., Spinal shock Spinal shock* Spinal shock refers to the period after spinal cord injury (usually 24-48 hours) when the reflex activity of the entirespinal cord becomes depressed. Spinal shock' occurs when the spinal cord fails temporarily following injury. During this period the reflex arcs below the level of spinal cord injury are notfunctioning. Spinal shock is basically flaccid paralysis with loss of all spinal reflexes. There is loss ofautonomic, reflex, motor and sensory activity below the level of the cord lesion.* Signs of spinal shock include:i) Flaccid paralysisii) Loss of deep tendon and perianal reflexes* The return of reflex activity below the level of a spinal cord injury signifies the end of spinal shock.* An accurate assessment of neurological status cannot be completed until spinal shock resolves. The earliest indicator of resolution is the return of bulbocavernous reflex. The significance of spinal shock lies in thedetermination of whether a patient has sustained a complete Vs incomplete spinal cord injury. This cannot be* determined until spinal shock has ended. If there is total absence of sensory or motor function below the level ofinjury, after spinal shock has resolved, It indicates complete cord injury. If there is some residual functiondistal to the level of injury after resolution of spinal shock indicates incomplete cord injury.* High thoracic cord has poor prognosis as this area has narrow spinal canal and precarious blood supply, watershadearea of blood supply lies in upper thoracic region.

So the answer to this question is A


Instruction:
What should be the density of microfilaria in blood to make it infective to the vector
A. 5 microfilaria / drop of blood
B. 10 microfilaria / drop of blood
C. 15 microfilaria / drop of blood
D. 20 microfilaria / drop of blood

Response:
A density of 15 microfilaria / drop of blood are needed to make it infective.

So the answer to this question is C


Instruction:
Beta-carbolin at benzodiazepine receptors acts as
A. Agonist
B. Antagonist
C. Inverse agonist
D. Physiological antagonist

Response:
Ans. c (Inverse agonist) (Ref. KDT 6th/ 395)INVERSE AGONISTS# An inverse agonist is an agent which binds to the same receptor binding-site as an agonist for that receptor but exerts the opposite pharmacological effect.# E.g.:- R015-4513 which is the inverse agonist of the benzodiazepine .- Gabapentin acts by facilitation of GABA-mediated CT channel opening. It does not block GABA receptors. Substances called beta-carboline which also bind this receptor cause stimulation anxiety, increased muscle tone and convulsions. They are inverse agonists of benzodiazepine receptors.Agonistsincrease the inhibitory transmitter GABA potencyInverse agonistsdecrease GABA potencyAntagonistsblock action of agonists. Drugs affecting GABAA-receptor gated chloride channel# GABA: Endogenous agonist at GABAA receptor - promotes Cl- influx# Buscimol: Agonist at GABAA site# Bicuculline: Competitive antagonist at GABAA receptor# Barbiturate: Agonist at an allosteric site (Ppicrotoxin site); prolong GABA action; open CI- channel# Alcohol, inhalational anaesthetics, Propofol: Open CI- channel directly; allosteric facilitation of GABA# Benzodiazepine: Agonist at an allosteric BZD site --^facilitate GABA action# b-Carboline (DMCM): Inverse agonist at BZD site ->>impede GABA action# Flumazenil: Competitive antagonist at BZD site

So the answer to this question is C


Instruction:
Why cereals and pulses are combined
A. 10% cereals contain protein and pulses contain 40%
B. Cereals are deficient in methionine and lysine is deficient in pulses
C. Cereals are deficient in lysine and methionine is deficient in pulses
D. Cereals are rich in essential AA

Response:
None

So the answer to this question is C


Instruction:
Which of the following is not a congenital anomaly?
A. Amastia
B. Polymastia
C. Polythelia
D. Mastitis

Response:
Mastitis is an inflammation of breast tissue that sometimes involves an infection. The inflammation results in breast pain, swelling, warmth and redness. You might also have fever and chills. Mastitis most commonly affects women who are breast-feeding (lactation mastitis) . Ref - medscape.com

So the answer to this question is D


Instruction:
Best laboratory test to diagnose HIV infection March 2007
A. ELISA
B. Western blot
C. Complement fixation test
D. RIA

Response:
Ans. B: Western blot

So the answer to this question is B


Instruction:
High plasma protein binding of a drug results in
A. Less drug interaction
B. High volume of distribution
C. Decreased glomerular filtration
D. Lowers duration of action

Response:
* Most drugs bind to plasma proteins in a reversible fashion without translating into a pharmacological response. (Plasma proteins are hence called Silent Receptors) * Acidic drugs bind to Albumin, Basic drugs to alpha1 acid- glycoprotein. * Binding of the drug to plasma protein affects the pharmacokinetic handling of drug by the body. Ref:- HL & KK Sharma's Principles of Pharmacology 3rd Ed; Pg Num:- 48

So the answer to this question is C


Instruction:
FTU is a measure of:
A. Area involved in severe drug reactions
B. Amount of topical drug to be applied
C. Mucosal involvement in pemphigus
D. Concentration of drug in topical preparations

Response:
An approximate practical measure of topical medication is the fingeip unit, the quantity of ointment, extruded from a tube with a nozzle of 5 mm diameter extending from the distal crease of the forefinger to ventral aspect of the fingeip. This unit weighs approximately 0.49 g in males and 0.43 g in females and covers, on average, an area of approx 300 cm2.

So the answer to this question is B


Instruction:
The FNAC of the lesion should reveal:
A. ‘Orphan-Annie eye’ nucleus cells
B. Amyloid deposits
C. Epitheloid cells and giant cells
D. Follicular cells

Response:
None

So the answer to this question is A


Instruction:
Endoscopic nasal surgery is indicated in-
A. Chronic sinusitis
B. Epistaxis
C. Both
D. None

Response:
None

So the answer to this question is C


Instruction:
Management of pancreatic abscess is
A. Needle aspiration
B. Cystogastrostomy
C. USG guided drainage
D. Cystojejunostomy

Response:
Sterile and Infected Pancreatic Fluid Collection (Pancreatic Abscess) The presence of acute abdominal fluid during an episode of AP has been described in 30-57% of patients In contrast to pseudocyst and cystic neoplasia of the pancreas, fluid collections are not surrounded or encased by epithelium or fibrotic capsule Treatment is suppoive because most fluid collections will be Spontaneously reabsorbed by the peritoneum Pancreatic Abscess: The presence of fever, elevated WBC count, and abdominal pain suggest infection of this fluid and percutaneous aspiration is confirmatory Percutaneous drainage and IV administration of antibiotics should be instituted if infection (Pancreatic abscess) is present Ref: Sabiston 20th edition Pgno :1529

So the answer to this question is A


Instruction:
Target material used for generating X-rays:
A. Tungsten
B. Cobalt
C. Cadmium
D. Palladium

Response:
Ans. Tungsten

So the answer to this question is A


Instruction:
Which of the following extrapyramidal effect is seen on chronic use of antipsychotic?
A. Dystonia
B. Akathisia
C. Tardive dyskinesia
D. Parkinsonism

Response:
Ans.C.Tardive dyskinesia Involuntary frowning, sticking out of the tongue, puckering, pursing, lip-smacking movements are seen in tardive dyskinesia due to chronic antipsychotic usage.Consider drug cessation or switch to an atypical antipsychotic in all patients.

So the answer to this question is C


Instruction:
Cardiac output peaks during pregnancy at what gestational age?
A. 20 weeks
B. 28 weeks
C. 32 weeks
D. 36 weeks

Response:
Ans. c (32 weeks). (Ref. Textbook of Obstetrics by Dutta, 4th ed., 53)Cardiac output starts to increase from 10th week of pregnancy, reaches its peak of 40% at about 24-30 weeks.PHYSIOLOGICAL CHANGES ASSOCIATED WITH PREGNANCYCardiovascular System1Blood Volume# Increases progressively from 6-8 weeks gestation (pregnancy)# Reaches a maximum at approximately 32-34 weeks.The increase in plasma volume (40-50%) is relatively greater than that of red cell mass (20-30%) resulting in hemodilution and a decrease in haemoglobin concentration.Intake of supplemental iron and folic acid is necessary to restore hemoglobin levels to normal (12 g/dl).The increased blood volume serves two purposes.First, it facilitates maternal and fetal exchanges of respiratory gases, nutrients and metabolites.Second, it reduces the impact of maternal blood loss at delivery.Typical losses of 300-500 ml for vaginal births and 750-1000 ml for Caesarean sections are thus compensated with the so-called "autotransfusion" of blood from the contracting uterus.2Blood ConstituentsAs mentioned above, red cell mass is increased by 20-30%. Leukocyte counts are variable during gestation, but usually remain within the upper limits of normal.Marked elevations, however, develop during and after parturition (delivery).3Fibrinogenas well as total body and plasma levels of factors VII, X and XII increase markedly.The number of platelets also rises, yet not above the upper limits of normal.Combined with a decrease in fibrinolytic activity, these changes tend to prevent excessive bleeding at delivery. Thus, pregnancy is a relatively hypercoagulable state, but during pregnancy neither clotting or bleeding times are abnormal.4Cardiac OutputIncreases to a similar degree as the blood volume.During the first trimester cardiac output is 30-40% higher than in the non-pregnant state.Steady rises are shown on Doppler echocardiography, from an average of 6.7 litres/ minute at 8-11 weeks to about 8.7 litres/minute flow at 36-39 weeks. They are due, primarily, to an increase in stroke volume (35%) and, to lesser extent, to a more rapid heart rate (15%).There is a steady reduction in systemic vascular resistance (SVR) which contributes towards the hyperdynamic circulation observed in pregnancy.5Cardiac Size/ Position/ECGThere are both size and position changes which can lead to changes in ECG appearance. The heart is enlarged by both chamber dilation and hypertrophy.

So the answer to this question is C


Instruction:
Which of the following causes of Anemia is associated with a Hypoplastic marrow -
A. Fanconi's Anemia
B. Paroxysmal Nocturanl Hemoglobinuria (PNH)
C. Hypersplenism
D. Myelofibrosis

Response:
None

So the answer to this question is A


Instruction:
The most accepted theory for dentin sensitivity is
or
Which of the following is the most accepted thoery
of transmission of dental sensation
A. Transduction theory
B. Direct neural stimulation
C. Hydrodynamic theory
D. Hydrostatic theory

Response:
None

So the answer to this question is C


Instruction:
Which of the following antibiotics acts by inhibiting cell wall synthesis?
A. Cefepime
B. Aminoglycosides
C. Erythromycin
D. Doxycycline

Response:
Ans. is 'a' i.e. Cefepime

So the answer to this question is A


Instruction:
Pseudo-Meigs syndrome syndrome seen with-
A. Fibroids
B. Adenomyosis
C. Endometriosis
D. Ovarian fibroma

Response:
Ans- A . Fibroids Meig's-ovarian fibromas Pseudomeig's-subserous Fibroid Pseudo-Meigs syndrome consists of pleural effusion (an example of which can be seen in the image below), ascites, and benign tumors of the ovary other than fibromas. These benign tumors include those of the fallopian tube or uterus and mature teratomas, struma ovarii, and ovarian leiomyomas.

So the answer to this question is A


Instruction:
All are true about short bowel syndrome except -
A. Intestinal transmit time is increased
B. Nutritional deficiency present
C. Site atorrhea present
D. None

Response:
None

So the answer to this question is A


Instruction:
Which of these antihypeensives do not have any central action?
A. Propranolol
B. Methyl dopa
C. Minoxidil
D. Nitroprusside

Response:
prpranolol is a beta blocker,methyldopa is a alpha agonist, minocidil is a potassium channel blocker and notroprusside is a nitrate all are used in treatment of hypeension. REF KD Tripathi 8th ed.

So the answer to this question is D


Instruction:
A woman presents to a dermatologist because she has lost almost all the hair on her body, including scalp hair, eyebrows, eyelashes, armpit and groin hair, and the fine hairs on her body and extremities. She most likely has a variant of which of the following?
A. Alopecia areata
B. Androgenic alopecia
C. Chronic cutaneous lupus erythematosus
D. Lichen planopilaris

Response:
Alopecia areata is caused by an autoimmune attack on hair follicles. It has a wide range of clinical severity, with most cases involving a localized patch of hair (which regrows within 1 year in half of the patients). The hair that does regrow may be gray or depigmented. More severe cases can involve the entire scalp (alopecia totalis) or, as in this patient, the entire body surface (alopecia universalis). These more severe cases are less likely to resolve adequately. Treatment of alopecia areata is often unsuccessful, but topical steroids are typically tried. Androgenic alopecia is common male pattern baldness. Chronic cutaneous lupus erythematosus can produce localized baldness. Lichen planopilaris can produce localized baldness. Ref: Landthaler M., Baumler W., Hohenleutner U. (2012). Chapter 239. Lasers and Flashlamps in Dermatology. In K. Wolff (Ed), Fitzpatrick's Dermatology in General Medicine, 8e.

So the answer to this question is A


Instruction:
Which of the following mullerian anomaly is least likely to develop endometriosis
A. Transverse vaginal septum
B. Bicornuate Uterus
C. Imperforate Hymen
D. Unicornuate uterus.

Response:
Bicornuate Uterus is least likely to develop endometriosis.

So the answer to this question is B


Instruction:
pANCA is sensitive and specified for-
A. Post streptococcal glomerulonephritis
B. Idiopathic cresentic glomerulonephritis
C. Diffuse glomerulosclerosis
D. None

Response:
None

So the answer to this question is B


Instruction:
A 39-year-old male presents with severe neck pain after a whiplash injury, sustained when his car was struck from behind. Radiographic studies reveal trauma to the ligament lying on the anterior surface of the cervical vertebral bodies. Which ligament is this?
A. Anterior longitudinal ligament
B. Ligamentum flavum
C. Nuchal ligament
D. Posterior longitudinal ligament

Response:
The anterior longitudinal ligament lies anterior to the vertebral bodies along the vertebral column. The ligamentum flavum connects the lamina of two adjacent vertebrae. The nuchal ligament is a continuation of the supraspinous ligament above C7, which connects spinous processes. The posterior longitudinal ligament lies on the posterior edge of the vertebral bodies. The transverse cervical (cardinal) ligament is associated with the pelvic region of the body and not the spinal column.

So the answer to this question is A


Instruction:
Impoant intermediate product of biosynthesis of fatty acid is
A. Cholesterol
B. Malonyl CoA
C. Acetyl CoA
D. Thioesterases

Response:
Carboxylation of Acetyl CoA The first step in the fatty acid synthesis is the carboxylation of acetyl CoA to form malonyl CoA. The acetyl CoA carboxylase is not a pa of the multienzyme complex. But it is the rate-limiting enzyme. Biotin, a member of B complex vitamins, is necessary for this reaction. The enzyme is allosterically regulated, the major effectors being citrate (positive) and palmitoyl CoA (negative). The reaction is similar to carboxylation of pyruvate to form oxaloacetate. The elongation of the fatty acid occurs by addition of 2 carbon atoms at a time. But the 2-carbon units are added as 3-carbon, malonyl units. The whole reaction sequence occurs while the intermediates are bound to ACP (acyl carrier protein).Ref: DM Vasudevan - Textbook of Biochemistry, 8th edition, page no: 137

So the answer to this question is B


Instruction:
Patient pays a stipulated amount of money towards the treatment is termed
A. Deductible
B. Coinsurance
C. Fee for service
D. Third party payment

Response:
None

So the answer to this question is A


Instruction:
Russel bodies are -
A. Intracellular accumulation of protein
B. Homogenous eosinophilic deposits
C. Seen in multiple myeloma
D. All of the above

Response:
None

So the answer to this question is D


Instruction:
Hepcidin associated with which molecule?
A. Iron
B. Copper
C. Selenium
D. Flurine

Response:
Ans. is 'a' i.e., Iron Hepcidin is a positive acute phase reactant. Hence during inflammation, liver releases more hepcidin. This inhibits basolateral side transpo of iron along intestine. This results in a hematological picture which mimics iron deficiency anemia. This is the basis of Anemia of Chronic Disease

So the answer to this question is A


Instruction:
A 19-year-old man with a history of recent-onset asthma presents with chest pain, intermittent claudication and respiratory distress that is unresponsive to bronchodilators and antibiotics. Physical examination reveals mild hypeension (blood pressure = 150/100 mm Hg), bilateral wheezing and numerous purpuric skin lesions on the feet. Laboratory studies demonstrate that leukocytes are increased to 14,000/mL with increased eosinophils and platelets are increased to 450,000/mL. BUN is elevated to 30 mg/dL, and serum creatinine is elevated to 3.5 mg/dL. The serum antineutrophil cytoplasmic antibody test is positive. Urinalysis discloses 3+ proteinuria and RBCs. A renal biopsy demonstrates vasculitis of medium-sized aeries, accompanied by eosinophilia. Which of the following is the most likely diagnosis?
A. Churg-Strauss disease
B. Henoch-Schonlein purpura
C. Loeffler syndrome
D. Wegener granulomatosis

Response:
Churg-Strauss disease is an idiopathic, systemic, granulomatous disease of small- and medium-size aeries characterized by vasculitis of many organs, fluctuating eosinophilia and late-onset asthma. The majority of patients display antineutrophil cytoplasmic antibodies. The disease is also known as allergic granulomatosis and angiitis. Transbronchial lung biopsy shows granulomatous lesions in vascular and extravascular sites, accompanied by intense eosinophilia. The vasculitis histologically resembles the lesions of polyaeritis nodosa and Wegener granulomatosis, but these diseases do not typically present with an asthmatic syndrome.

So the answer to this question is A


Instruction:
Rytand's murmur is heard in
A. Mitral Stenosis
B. PDA
C. Complete AV block
D. Acute rheumatic fever

Response:
Rytand's murmur is mid diastolic murmur in complete heart block.

So the answer to this question is C


Instruction:
Necrosis of cells is due to :
A. Ca2+ efflux cell
B. Fat deposition in cells.
C. Water imbibition in cells.
D. Enzymatic digestion.

Response:
• Necrosis: It refers to morphological changes that follow cell death in living tissues.
• These morphological changes are the result of 2 concurrent processes.
1. Enzymatic digestion.
2. Denaturation of proteins.
• The enzymes used in the process are derived from 2 sources:
1. From the lysosomes of the dead cells themselves (Autolysis).
2. From the lysosomes of immigrant leucocytes (Heterolysis).

So the answer to this question is D


Instruction:
Foot drop due to injury of which nerve
A. Tibial nerve
B. Deep peroneal nerve
C. Superficial peroneal nerve
D. Sural nerve

Response:
FOOT DROP:- Paralysis of muscles of anterior compament of the leg due to injury to deep peroneal nerve results in loss of power of dorsiflexion of foot and eversion. As a result foot us plantar flexed. Ankle reflex is intact.This is usually caused by injury or disease of the common peroneal nerve due to trauma, leprosy or peripheral neuritis.Sensory loss- first interdigital cleft. {Reference: BDC 6E pg no.97}

So the answer to this question is B


Instruction:
Myocardial  contractility  is  increased by:
A. Atropine
B. Decreased end diastolic volume
C. Increased  heart  rate  from  70  to  150 beats/min
D. Reduced arterial pH to 7.3

Response:
Option A: Atropine is a muscarinic receptor antagonist that is used to inhibit the effects of excessive vagal activation on the heart, which is manifested as sinus bradycardia and AV nodal block. Effect on contractility is negligible.
Option B: Decrease EDV decreases contractility by Frank Starling law.
Option  D:  Acidosis, hypercapnoea and hypoxia inhibit myocardium contractility.
Option C: This effect is known as Bowditch Effect (Myocardial tension increases with an increase in heart rate).

So the answer to this question is C


Instruction:
Premalignant conditions of stomach include all, Except:
A. Gastric ulcer
B. Pernicious anemia and achlorhydria
C. Hiatus Hernia
D. Atrophic gastritis

Response:
Answer is C (Hiatus Hernia): Hiatus Hernia is not considered a premalignant lesion for gastric carcinoma. Premalignant lesions for Gastric Adenocarcinoma Chronic Atrophic Gastritis (H. pylori > Pernicious Anemia) Intestinal Metaplasia Type II and Type III (Intestinal Metaplasia type 1 is not associated with increased risk) Gastric Dysplasia Gastric polyps (Adenomatous polyps) (Hyperplastic polyps are generally not considered premalignant as the rate of malignant transformation in these polyps is <1% and confined to polyps > lcm) Previous Gastrectomy Menetrier's Disease Gastric ulcers (long standing gastric ulcers : Low risk)

So the answer to this question is C


Instruction:
Genetic disorders are diagnosed at 11 weeks of pregnancy by :
A. Chorionic villous biopsy
B. Amniocentesis
C. Endometrial biopsy
D. Placentography

Response:
Chorionic villous biopsy

So the answer to this question is A


Instruction:
All of the following are immune complex diseases except-
A. Serum sickness
B. Farmer's lungs
C. SLE
D. Graft rejection

Response:
Ans. is 'd' i.e., Graft rejectionGraft rejection is Type IV hypersensitivity

So the answer to this question is D


Instruction:
Most important layer of a slow sand filter is -
A. Vital layer
B. Sand bed
C. Filter system
D. Raw water

Response:
This is confusing one because of the following statements of Park.
	"The most important part of the filter is the sand bed" - Park
	"The vital layer is the heart of the slow sand filter" - Park
Following statement may help in choosing the answer : -

"Vital layer is the layer that provides the effective purification in potable water treatment, the underlying sand layer providing the support medium for this biological treatment layer"  - Water & Health 3rd/e 733.

Read the mechanism of action of slow sand filter and there will be no confusion.
Mechanism of action of slow sand filter
Slow sand filter work through the, formation of a gelatinous layer called vital layer or hypogeal layer or Schmutzdecke in the top few cm of sand layer.
This layer is formed in the first 10-20 days of operation.
The formation of vital layer is known as "Ripening of the filter".
This vital layer consists of bacteria, fungi, Protozoa, and a range of aquatic insect larvae.
As waste passes through vital layer, particles of foreign matter are trapped in the mucilaginous matrix and dissolved organic material is absorbed and metabolized by bacterial, fungi and protozoa → Vital layer has mechanical (physical) as well as biological action.
So, the vital layer acts as the heart of slow sand filter.

So the answer to this question is A


Instruction:
ETDRS vision chart study is done in patients with diabetic retinopathy. ETDRS stands for –
A. Extended therapy for diabetic retinopathy & its sutdy
B. Emergency treatment for diabetic retinopathy and study
C. Eye testing or rotatory drum & its study
D. Early treatment for diabetic retinopathy study

Response:
Early treatment diabetic retinopathy study (ETDRS) was a multicenter, randomized clinical trial involving 3711 participants designed to evaluate the effectiveness of both organ laser photocoagulation and aspirin therapy in the management of patients with non-proliferative diabetic retinopathy and early proliferative diabetic retinopathy.
In addition, it was designed to determine the best time to initiate photocoagulation treatment in diabetic retinopathy.

So the answer to this question is D


Instruction:
Sunflower type cataract is characteristically seen in-
A. Chalcosis
B. Diabetes
C. Stragardt's disease
D. Congenital syphilis

Response:
Ans. is 'a' i.e., Chalcosis o Sunflower cataract is seen in Wilson's disease, chalcosis (copper deposition), penetrating trauma with retention of copper or its alloy.DiseaseType of cataracto Myotonic dystrophyChristmas tree cataract, Blue dot cortical cataract, posterior subcapsular cataract.o Wilson's diseaseo Copper deposition (Chalcosis)o Penetrating traumaSunflower cataract (Petals of flower cataract)o Diabetes mellitusSnowflake cataract or Snowstrom cataracto Down's syndromeSnowflake cataracto Hypoparathyroidism (Hypocalcemia) Hypovitaminosis DDiscrete white cortical opacities, Congenital lamellar (zonular) cataracto Atopic dermatitisBlue dot cortical cataract, posterior subcapsular cataracto Congenital rubellaNuclear cataract, Total congenital cataracto GalactosemiaOil drop cataract, Dust like lenticular opacitieso Complicated cataract (Iridocyclitis, High myopia)Posterior cortical breads's crumb appearance. Polychromatic lusture.Rainbow cataracto Blunt traumaVossius ring on anterior surface of lens. Rosette shaped cataract

So the answer to this question is A


Instruction:
In PCR Acquaticus thermophilus is preferred over E.coli. because -
A. Thermostable at temperature at which DNA liquefies
B. Proofreading done
C. Done in more precisely
D. Does not require primer

Response:
None

So the answer to this question is A


Instruction:
Ketone body formation without glycosuria is seen in ?
A. Diabetes mellitus
B. Diabetes insipidus
C. Starvation
D. Obesity

Response:
Ans. is 'c' i.e., StarvationAmongst the given options, DM and starvation are the causes of ketosisDiabetes :- Ketosis with hyperglycemia and glycosuriaProlonged starvation :- Ketosis with low or normal glucose and without glycosuria.In diabetic Ketoacidosis:- (i) Positive Rothera's test (due to ketone bodies) (ii) Positive Benedict's test (due to presence of reducing sugar in urine)In Starvation ketosis:- (i) Postive Rothera's test (due to ketone bodies), (ii) Negative Benedict's test (no sugar in urine)

So the answer to this question is C


Instruction:
Paraneoplastic opsoclonus myoclonus syndrome is most commonly associated with which malignancy in children?
A. Hepatoblastoma
B. Hepatocellular carcinoma
C. Lymphoma
D. Neuroblastoma

Response:
Opsoclonus is a disorder of eye movement characterized by involuntary, chaotic saccades that occur in all directions of gaze; it is frequently associated with myoclonus and ataxia. Opsoclonus-myoclonus may be cancer-related or idiopathic. When the cause is paraneoplastic, the tumors involved are usually cancer of the lung and breast in adults and neuroblastoma in children. Ref: Harrison's principle of internal medicine 17th edition, chapter 97.

So the answer to this question is D


Instruction:
The paradoxical response of GH release to TRH is seen in -
A. Prolactinoma
B. Acromegaly
C. Malnutrition
D. Pitutary adenoma

Response:
TRH stimulates the secretion of prolactin as well as TSH,
Normally it does not affect growth hormone secretion but in acromegaly TRH induced growth hormonal stimulation Occurs.

Note:
"In prolactinomas paradoxical response of TRH to prolactin occurs".
In normal individuals,
TRH stimulates prolactin release
Where as in prolactinomas
There is little or no rise in prolactin in response to TRH.

So the answer to this question is B


Instruction:
Ptosis results from trauma to which nerve
A. VII
B. VIII
C. III
D. VI

Response:
Ans. is 'c' i.e. III C.N. Palsy of Levator palpebrae superioris supplied by III C.N. results in ptosis.

So the answer to this question is C


Instruction:
Filters of which metal that is used to remove long wave
length X-rays from the primary beam is:
A. Platinum
B. Aluminium
C. Copper
D. Lead

Response:
None

So the answer to this question is B


Instruction:
Mechanism of direct transfer of free DNA-
A. Transformation
B. Conjugation
C. Transduction
D. None

Response:
None

So the answer to this question is A


Instruction:
Which of the following is primarily a venodilator ?
A. Hydralazine
B. Minoxidil
C. Nitroprusside
D. Nifedipine

Response:
Ans. is 'c' i.e., Nitroprusside o Hydralazine, minoxidil and nifedipine are primarily aeriolar dilators whereas nitroprusside is a mixed aeriolar and venous dilator, therefore best answer here seems to be nitroprusside.

So the answer to this question is C


Instruction:
A 25-year-old female presented in the month of December with chronic fatigue, cyanosis with bluish lips and ahralgia. Peripheral blood film is shown below. What is the likely cause?
A. Cold AIHA
B. Warm AIHA
C. Hemoglobinopathy
D. G6PD Deficiency

Response:
The given figure is of RBC agglutination. The history is suggestive of agglutination occurring in the month of December that is cold induced. Cold Agglutinin AIHA immune hemolytic anemia caused by IgM antibodies that bind red cells avidly at low temperatures (0deg-4degC). antibodies appear transiently following ceain infections, such as with Mycoplasma pneumoniae, EBV, CMV, influenza virus, and HIV. Clinical symptoms result from binding of IgM to red cells in vascular beds where the temperature may fall below 30degC, such as in exposed fingers, toes, and ears.

So the answer to this question is A


Instruction:
Dental papilla gives rise to
A. Pulp only
B. Pulp & Dentin
C. Pulp & periodontal ligament
D. All of the above

Response:
The cells of dental papilla will form pulp and dentin and the cells of dental sac will form cementum and periodontal ligament.

So the answer to this question is B


Instruction:
Autonomic nervous system the cholinergic nerve fibres
A. All preganglionic fibres, releases acetycholine
B. Post ganglionic cholinergic fibres secretes acetyl choline
C. The post ganglionic sympathetic cholinergic nerve fibres supplies sweat glands and blood vessels inheart and skeletal muscles
D. All of the above

Response:
(All of the above) (223 - 24-G 22nd) (266-68- Ganong 23rd)Neurotransmitters of ANSSympathetic fibres1. Preganglionic fibres - Acetylcholine2. Post ganglionic nor adrenergic fibres - Acetylcholine3. Post ganglionic cholinergic nerve fibres - AcetylcholineThe post ganglionic sympathetic cholinergic nerve fibres supply sweat glands blood vessels in heart and in skeletal musclesParasympathetic fibres1. Preganglionic fibres - Acetylcholine2. Post ganglionic fibres - Acetyl choline

So the answer to this question is D


Instruction:
Most common neonatal disorder screened is:
A. Neonatal hypothyroidism
B. Neonatal hypehyroidism
C. Hemoglobinopathies
D. Congenital Dislocation of Hip

Response:
Most common neonatal disorder to be screened is Neonatal hypothyroidism (NNH) Blood sample is collected from Cord's Blood /fromheel prick after 24hrs of bih Test- measurement of T4 or TSH /both simultaneously. As a single method, T4 is more useful (greater precision and reproducibility) Congenital Hypothyroidism is one of the most common preventable cause of mental retardation. Hence, neonatal screening & early supplementation of thyroid hormones can prevent this mental retardation.

So the answer to this question is A


Instruction:
Suburethral metastasis is seen in :
A. Carcinoma cervix
B. Carcinoma endometrium
C. Choriocarcinoma
D. Vaginal carcinoma

Response:
Carcinoma endometrium

So the answer to this question is B


Instruction:
Nihilistic ideas are seen in -a) Simple schizophreniab) Paranoid schizophreniac) Cotard's syndromed) Depressione) Body dysmorphic disorder
A. cd
B. bc
C. bd
D. ac

Response:
Nihilistic delusion is false belief that oneself, other or the world is non-existent or about to end. It is seen in severe depression.
If nihilistic delusions are accompanied by ideas concerning bodily function failure, the syndrome is called Cotard's syndrome.

So the answer to this question is A


Instruction:
A new test- test B for Diabetes mellitus is performed in a group containing diabetics and non-diabetics.A standard test A has been used as reference to classify them as diabetics and non diabetics. If 120 mg/dl of blood sugar is taken as the cut-off for test B, the area shaded in the graph below represents -
A. True Positives
B. False Positives
C. True Negatives
D. False Negatives

Response:
Ans- The shaded area indicated has Diabetics proved by a standard test A with blood sugar levels less than 120 mg/dl. Which is less than the cut-off as per the test B criteria required to label them as Diabetic . Hence, even though the shaded area has diabetics - but not picked up by this particular test B- known as False Negative.

So the answer to this question is D


Instruction:
Beer - claw appearance on CECT abdomen is seen in
A. Hepatic laceration
B. Pancreatic laceration
C. HCC
D. RCC

Response:
Beer-claw laceration : Multiple linear laceration of liver on CECT abdomen Ref: Sabiston 20th edition Pgno : 437-439

So the answer to this question is A


Instruction:
Increased airway resistance is/are caused by
A. Forced expiration
B. Dense air
C. Low lung volume
D. All

Response:
A, B, C i.e. Forced expiration, Dense air, Low lung volume

So the answer to this question is D


Instruction:
Periodic acid schiff stain shows Block positivity
A. Myeloblasts
B. Lymphoblasts
C. Monoblasts
D. Megakaryoblasts

Response:
Lymphoblast contain cytoplasmic aggregates of periodic acid-schiff (PAS)-positive material.

So the answer to this question is B


Instruction:
In the given condition, cause of defective vision is
A. Astigmatism
B. Hypermetropia
C. Visual axis is obscured
D. Diplopia

Response:
Pterygium: - Definition:- A degenerative condition of the subconjunctival tissue which proliferates as vascularised granulation tissue to invade the cornea . Pathogenesis:- * Elastotic degeneration of the subconjunctival tissue with Fibrovascular proliferation . * Destruction of epithelium & Bowman's layer by advancing fibrovascular tissue resulting in corneal scarring. Cause Of Defective Vision In Pterygium:- * Fibrovascular tissue contracts --> flattening of the cornea in the horizontal axis --> ASTIGMATISM {WITH THE RULE ASTIGMATISM } * Obscuring the pupillary region or the Visual axis. Types Of Pterygium:- Depending upon the growth * STOCKER'S LINE:- seen in atrophic pterygium. Iron deposition in front of the head in the corneal epithelium. Ref:- Parson's Diseases of the eye 22nd edition; pg num:- 184

So the answer to this question is C


Instruction:
Patients with mutations leading to defective beta-oxidation of fatty acid usually have hypoglycemia because of
A. Less ATP for gluconeogenesis
B. Less acetyl CoA for glucose formation
C. Increased fatty acids decrease glucose utilization
D. Fatty acids inhibit transpo of glucose for metabolism

Response:
Acetyl-coenzyme A or acetyl-CoA is an impoant molecule that paicipates in many biochemical reactions in protein, carbohydrate, and lipid metabolism. Its main function is to deliver the acetyl group to the citric acid cycle (Krebs cycle) to be oxidized for energy production. Ref-Sathyanarayana 4/e p399

So the answer to this question is B


Instruction:
Life style modification (discouraging children from adopting harmful lifestyles) is: March 2013 (b)
A. Primary level of prevention
B. Secondary level of prevention
C. Teiary level of prevention
D. Primordial prevention

Response:
Ans. D i.e. Primordial prevention Primordial prevention is defined as prevention of the development of risk factors before they happen.

So the answer to this question is D


Instruction:
All of the following may lead to plaque formation except
A. Pityriasis rosea
B. Psoriasis
C. Pemphigus
D. Lichen planus

Response:
Pemphigus vulgaris is a immunobullous disorder characterised by formation of flaccid blisters in skin and mucous membranes. other options lesions are characterised by papules and plaques. iadvl textbook of dermatology page 1088

So the answer to this question is C


Instruction:
Birbeck granules in the cytoplasm are seen in-
A. Mast cells
B. Langerhan's cells
C. Thrombocytes
D. Myelocytes

Response:
Birbeck granules, also known as Birbeck bodies, are rod shaped or "tennis-racket" cytoplasmic organelles with a central linear density and a striated appearance. First described in 1961 (where they were simply termed "characteristic granules"), they are solely found in Langerhans cells. Although pa of normal Langerhans cell histology, they also provide a mechanism to differentiate Langerhans cell histiocytoses (which are a group of rare conditions collectively known as histiocytoses) from proliferative disorders caused by other cell lines. Formation is induced by langerin.

So the answer to this question is B


Instruction:
In DIC, which is/are seen -
A. Normal APTT
B. Increased PT
C. Increased factor VIII
D. None

Response:
Ans. is 'b' i.e., Increased PT; Laboratory findings in DIC o Low platelet count. o Blood film shows microangiopathic haemolytic anaemia. There is presence of fragmented R.B.C. due to damage caused by trapping and passage through fibrin thrombi. o PT, thrombin time, and Activated Paial thromboplastin time all are prolonged. o Plasma fibrinogen level decreased. o Fibrin degradation product (FDPS) are raised. o Factor V and factor VIII decreased in concentration.

So the answer to this question is B


Instruction:
Gestational sac is seen by TVS at-
A. 5 weeks
B. 6 weeks
C. 7 weeks
D. 8 weeks

Response:
Ans. is 'a' i.e., 5 weeks [Ref: Dutta's Obs 7th/ep. 68}TVS showso Gestational sac & yolk sac - 5th weeko Fetal pole & cardiac activity - 6th weeko Embryonic movements - 7th weeko Fetal gestational age is best determined by CRL between 7th to 12th weeks (variation +/- 5 days)o Doppler effect of ultrasound can pick up heart rate by 10lh week.

So the answer to this question is A


Instruction:
Which tumour is not seen in the anterior mediastinum -
A. Teratoma
B. Thymic tumours
C. Thyroid tumours
D. Neurofibroma

Response:
Ans. is `d' i.e., Neurofibroma Mediastinal Tumors and Other Masses Superior Mediastinum Anterior Mediastinum Posterior Mediastinum Middle Mediastinum Lymphoma Thymoma Neurogenic tumors Bronchogenic cyst Thymoma Teratoma Lymphoma Pericardial cyst Thyroid lesions Lymphoma Gastroenteric hernia Lymphoma Metastatic carcinoma Thyroid lesions Parathyroid tumors Parathyroid tumors

So the answer to this question is D


Instruction:
A patient with a previous history of tuberculosis now complains of hemoptysis. Chest x-ray reveals an upper lobe mass with a cavity and a crescent-shaped air-fluid level. Select the most likely etiologic agent.
A. Aspergillus flavus
B. Coccidioides immitis
C. Herpes simplex virus type 1
D. Herpes simplex virus type 2

Response:
The patient who has a history of tuberculosis and now complains of haemoptysis would be revaluated for active tuberculosis. However, the chest x-ray described is characteristic of a fungus ball--almost always the result of an aspergilloma growing in a previous cavitary lesion.

So the answer to this question is A


Instruction:
Rao and Singh's index is?
A. Age-dependent index
B. Age-independent index
C. Age and sex-dependent index
D. Independent of wt and ht

Response:
RAO & SINGH'S INDEX: - It is calculated as Weight (in kg) x 100/height2 (in cm) - Normal value: 0.14 - In children with malnutrition: 0.12 - 0.14

So the answer to this question is B


Instruction:
Major source of estrogen in postmenopausal women is
A. Peripheral conversion of androstenedione
B. Secretion from adrenals
C. Testosterone from the ovary
D. H

Response:
All postmenopausal and symptomatic per menopausal women who do not have contraindication paicularlyThose with high-risk factors osteoporosis.Those with premature menopauseEvaluation before staing H:History-personal, medical, gynecological, familyExamination -HT, wt., BMI, BP, breast, vaginal examinationPap smearMammographyLipid profileTransvaginal sonography-endometrial thicknessWork up relevant to symptoms, e.g. urinary continence, carbohydrate intolerance, psychoemotional states.Contraindications of H:PregnancyCA breastCA endometriumUndiagnosed abnormal vaginal bleedingActive liver diseasePresent deep vein thrombosis, thromboembolic disorderInvestigation not routinely needed before staing HEndometrial sampling-Only did if the history of abnormal vaginal bleeding, TVS shows endometrial thickness> 5 mm.Serum FSH-Only did if the diagnosis of menopause is not clear, e.g. vasomotor symptoms in absence of amenorrhoea Bone densitometry-Using DEXA (dual energy X-ray absorptiometry) of lumber veebrae, hip bone for women with risk factors for osteoporosis, persistent backache, bone aches, history of recent fracture, for research purpose

So the answer to this question is A


Instruction:
Obturation may be done in the initial visit if
A. There is necrotic pulp
B. There is a swelling
C. The tooth is vital
D. The tooth is nonvital

Response:
None

So the answer to this question is C


Instruction:
Stress-induced hyperglycemia is due to:
A. Glucocoicoids
B. GH
C. Epinephrine
D. All

Response:
A i.e. Glucocoicoids; B i.e. GH; C i.e. Epinephrine

So the answer to this question is D


Instruction:
Heavy pressure application during mask ventilation may compress:
A. Marginal mandibular nerve.
B. Laryngeal nerve.
C. Arytenoid cartilage.
D. All of the above.

Response:
Heavy pressure application doing mask ventilation may compress the marginal – mandibular branch of facial N in angle of mandible Region.

So the answer to this question is A


Instruction:
Gluten sensitive enteropathy is most strongly associated with which of the following?
A. HLA DQ2
B. HLA DR4
C. HLA DQ3
D. Blood group 'B'

Response:
95% of patients with gluten sensitive enteropathy express HLA DQ2 histocompatibility antigen on chromosome 6. In celiac sprue, the fundamental disorder is a sensitivity to gluten. When small intestinal mucosa is exposed to gluten, there is intraepithelial accumulation of cytotoxic T cells and helper T cell in the lamina propria. Cytokines released by the T cells in turn damage intestinal enterocytes.Ref: Robbins Pathologic Basis of Disease, 6th Edition, Page 813.

So the answer to this question is A


Instruction:
Prophylaxis of is essential to prevent which vitamin deficiency in patients taking isoniazid?
A. B1
B. B2
C. B3
D. B6

Response:
Ans. is 'd' i.e., B6 * Isoniazid reacts with pyridoxal to form a hydrazone, and thus inhibits gneration of pyridoxal phosphate.* Isoniazid also combines with pyridoxal phosphate to interfere with its coenzyme function. Due to formation of hydrazones, the renal excretion of pyridoxine compounds is increased.* Thus, isoniazid therapy produces a pyridoxine deficiency state.

So the answer to this question is D


Instruction:
Huhle cell carcinoma is a variant of ?
A. Medullary carcinoma
B. Papillary carcinoma
C. Follicular carcinoma
D. Anaplastic carcinoma

Response:
Ans. is 'c' i.e., Follicular carcinoma Follicular carcinoma Second most common type of thyroid cancer. o Huhle cells are seen. o Differentiated from follicular adenoma by capsular and/or vascular invasion. o Unlike in papillary carcinoma, lymphatic spread is rare, and vascular invasion is common in follicular cancers. o Huhle cell or oncocytic carcinoma is a varriant of follicular carcinoma.

So the answer to this question is C


Instruction:
Retentive grooves in the proximal box of a class II amalgam cavity should be
A. Sharp and elongated at the dentinoenamel junction
B. Round and elongated in the dentin
C. Round and short in the dentin
D. Sharp and elongated in the dentin

Response:
None

So the answer to this question is B


Instruction:
A 30-year-old lady presents with features of malabsorption and iron deficiency anemia. Duodenal biopsy shows complete villous atrophy. Which of the following antibody is likely to be present?
A. Antiendomysial antibodies
B. Anti-goblet cell antibodies
C. Anti-Saccharomyces cerevisiae antibodies
D. Antineutrophil cytoplasmic antibodies

Response:
None

So the answer to this question is A


Instruction:
Drug of choice in methicillin-resistant Staphylococcus aureus -
A. Ceftriaxone
B. Ceftazidime
C. Ampicillin
D. Vancomycin

Response:
Methicillin resistance in S. aureus is determined by the mecA gene, which encodes for resistance to all b-lactam drugs. If CA-MRSA infection is suspected, either linezolid or vancomycin should be added to the initial empirical regimen. (Harrison's Principles of internal medicine, 20th edition)

So the answer to this question is D


Instruction:
Most common site of subphrenic abscess is
A. Right anterior intraperitoneal space
B. Right posterior intraperitoneal space
C. Left anterior intraperitoneal space
D. Left posterior intraperitoneal space

Response:
Right posterior intra-peritoneal space(Right subhepatic space/Rutherford Morison's kidney pouch) is bounded in front by the liver & gallbladder,above by the liver,behind by the right kidney & diaphragm,below by the transverse colon & hepatic flexure,to the left by foramen of Winslow & duodenum.It is the largest and deepest space of all.It is the most common site of subphrenic abscess.Causes include appendicitis,cholecystitis,post operative,perforated duodenal ulcer,intestinal obstruction. Reference:SRB' s manual of surgery,5th edition,page no:575.

So the answer to this question is B


Instruction:
Inveed fir tree appearance is characteristic of
A. Bacillus anthrax
B. Hemophilus influenzae
C. Yersinia pestis
D. Brucella

Response:
Anthrax bacilli is gram-positive and is found singly, in pairs or in sho chains. On gelatin stab culture, a characteristic inveed fir tree appearance is seen in the case of anthrax bacilli, with slow liquefaction commencing from the top. On blood agar, the colonies are non-hemolytic, though occasional strains produce a narrow range of hemolysis. In broth, growth occurs as floccular deposits, with little or no turbidity. Reference: Textbook of Microbiology; Anathanarayan and panicker's; 10th edition; Page no: 251

So the answer to this question is A


Instruction:
What is the inheritance pattern of friedreich ataxia?
A. Autosomal dominant
B. Autosomal recessive
C. x linked dominant
D. x linked recessive

Response:
Friedreich Ataxia and Ataxia Telanjectasis both are autosomal recessive disorder.

So the answer to this question is B


Instruction:
Glucogenic amino acid is-
A. Leucine
B. Lysine
C. Glutamine
D. None

Response:
Ans. is 'c' i.e.. Glutamine GlucogenicKetogenicBoth glucogenic & ketogenico Alanineo Arginineo Aspargineo Aspartic acido Cysteineo Glutamineo Glutamic acido Glycineo Histidineo Methionineo Prolineo Serineo Threonineo Valineo Leucineo Lysineo Isoleucineo Phenylalaineo Tyrosineo TryptophanNote : According to Harper 29PSh/'e (P-282), lysine is both ketogenic and glucogenic. But all other textbooks have mentioned it as ketogenic.

So the answer to this question is C


Instruction:
A 34-year-old power lifter visits the outpatient clinic because he has difficulty walking. During physical examination it is observed that the patient has a problem unlocking the knee joint to permit flexion of the leg. Which of the following muscles is most likely damaged?
A. Biceps femoris
B. Gastrocnemius
C. Popliteus
D. Semimembranosus

Response:
When the popliteus contracts, it rotates the distal portion of the femur in a lateral direction. It also draws the lateral meniscus posteriorly, thereby protecting this cartilage as the distal femoral condyle glides and rolls backward, as the knee is flexed. This allows the knee to flex and therefore serves in unlocking the knee. The biceps femoris is a strong flexor of the leg and laterally rotates the knee when it is in a position of flexion. The gastrocnemius is a powerful plantar flexor of the foot. The semimembranous, similar to the biceps femoris, is a component of the hamstring muscles and is involved in extending the thigh and flexing the leg at the knee joint. The rectus femoris is the strongest quadriceps muscle in extending the leg at the knee.

So the answer to this question is C


Instruction:
40 year old lady presented with sudden onset of breathing difficulty. Clinical examination was essentially normal except for tachycardia and tachypnea. She gives history of two episodes of lower limb pain and swelling, first episode was 15 years back during her first pregnancy and second episode 3 months back. What would be the hematological abnormality respoinsible for her condition:
A. Protein C excess
B. Resistance to activated protein C
C. Antithrombin mutation causing accelerated effect
D. Polycythemia vera

Response:
Here the patient is probably suffering from pulmonary embolism as the major symptom of pulmonary embolism is sudden onset of breathlessness. Chest examination will be normal. As the patient gives history of recurrent episodes of lower limb swelling, she must be suffering from reccurrent deep vein thombosis as a result of some thrombophilic condition. FACTS: Heritable causes of venous thrombosis 1. Antithrombin deficiency and factor V leiden mutation 2. Protein C and S deficiency 3. Activated protein C resistance 4. Polycythemia can cause venous thrombosis but in this patient there are no features to suggest polycythema Ref: Harrison, Edition - 18, Page - 987

So the answer to this question is B


Instruction:
A 20 years old female has thoughts of cutting her fingers, she plans and imagines doing it but never actually does it. She says that she is not having any guilt of having such thought. And also says the thoughts are distressing her and she is unable to control them. The thoughts vanish either by ending with a seizure or automatically subsides on its own. Which of the following is the cause?
A. Thought inseion
B. Obsession
C. Forced thinking
D. Thought crowding

Response:
Answer- C. Forced thinkingForced thinking is a rare epileptic phenomenon, usually seen in patients with frontal lobe epilepsy. It is a rare type of aura that refers to recurrent intrusive thoughts, ideas, or crowding of thoughts. The above profile fits into the criteria for epileptic forced thinking, which is a rare phenomenon seen in patients with frontal lobe epilepsy.

So the answer to this question is C