Instruction:
An adult patient presents with normal or enlarged kidneys with massive proteinuria. Most likely cause is -
A. Chronic pyelonephritis
B. Chronic glomerulonephritis
C. Amyloidosis
D. Renal aery stenosis

Response:
Answer- C. AmyloidosisAmyloidosis causes :-Enlarged kidneyMassive proteinuria (nephrotic range proteinuria)

So the answer to this question is C


Instruction:
Features not seen in Cushing&;s Syndrome
A. Hypoglycemia
B. Hypeension
C. Frank psychosis
D. Hypokalemia

Response:
In Cushing&;s syndrome, hyperglycemia usually occur; hyperglycemia may progress to frank diabetes. Page 1940,Nelson textbook of pediatrics 19th edition

So the answer to this question is A


Instruction:
The tensile strength of wound reaches that of tissue by ?
A. 6 weeks
B. 2 months
C. 4 months
D. None

Response:
Ans. is 'None' It never equals the tensile strength of unwounded skin. Robbin's writes- "How long does it take for a skin wound to achieve its maximal strength? When sutures are removed from an incisional surgical wound, usually at the end of the first week, wound strength is approximately 10% that of unwounded skin. Wound strength increases rapidly over the next 4 weeks, slows down at approximately the third month after the original incision, and reaches a plateau at about 70% to 80% of the tensile strength of unwounded skin."

So the answer to this question is D


Instruction:
Normal sperm concentration in semen analysis should be
A. >= 5 millilon/ml
B. >= 15 millilon/ml
C. >= 20 millilon/ml
D. >= 25 millilon/ml

Response:
Normal Semen Analysis: World Health Organization Volume>=1.5 mLSperm concentration>=15 million/mL Sperm motility>=32% progressive>25% rapidly progressiveMorphology (strict criteria)>=4% normal formWhite blood cells<1 million/mLReference: Novak's Gynecology; 14th edition; Chapter 32; Infeility and Assisted Reproductive Technology

So the answer to this question is B


Instruction:
Sorghum contains excess of ?
A. Lycine
B. Arginine
C. Leucine
D. Histidine

Response:
Ans. is 'c' i.e., Leucine Jowar (Sorghum) contains excess of leucine which can cause niacine deficiency and pellagra.

So the answer to this question is C


Instruction:
Friction locked pins are usually retained by
A. Threads on pin
B. Resiliency of dentine
C. Luting cement
D. Channel depth

Response:
None

So the answer to this question is B


Instruction:
Inhibition of bile secretion will NOT reduce the intestinal absorption of:
A. Vitamin A
B. Vitamin B1
C. Vitamin K
D. Vitamin D

Response:
Inhibition of bile secretion will have little effect on the intestinal absorption of a water-soluble vitamin B1 (thiamine). All of the other structures listed are fat-soluble and for a successful intestinal absorption should be emulsified by the bile acids (glycocholic and taurocholic acid). Ref: Textbook Of Biochemistry And Human Biology, 3rd Ed, Page 524

So the answer to this question is B


Instruction:
Painless effusion in joints in Congenital syphilis are called
A. Clutton's joint
B. Charcot's joint
C. Barton's joint
D. Chronic osteomyelitis

Response:
Painless synovitis at puberty in late congenital syphilis is called clutton's joint.

So the answer to this question is A


Instruction:
Features of hypovolemic shock are all except -
A. Oliguria
B. Bradycardia
C. Low B.P
D. Acidosis

Response:
None

So the answer to this question is B


Instruction:
Several weeks after surgical dissection of her left axilla for the removal of lymph nodes for staging and treatment of her breast cancer, a 32-year-old woman was told by her general physician that she had "winging" of her left scapula when she pushed against resistance during her physical examination. She told the physician that she had also experienced difficulty lately in raising her right arm above her head when she was combing her hair. In a subsequent consult visit with her surgeon, she was told that a nerve was accidentally injured during the diagnostic surgical procedure and that this produced her scapular abnormality and inability to raise her arm normally. What was the origin of this nerve?
A. The upper trunk of her brachial plexus
B. The posterior division of the middle trunk
C. Roots of the brachial plexus
D. The posterior cord of the brachial plexus

Response:
The long thoracic nerve was injured during the axillary dissection, resulting in paralysis of the serratus anterior. The serratus anterior is important in rotation of the scapula in raising the arm above the level of the shoulder. Its loss results in protrusion of the inferior angle ("winging" of the scapula), which is more obvious when one pushes against resistance. The long thoracic nerve arises from brachial plexus roots C5, C6, and C7. The upper trunk (C5,C6) supplies rotator and abductor muscles of the shoulder and elbow flexors. The posterior division of the middle trunk contains C7 fibers for distribution to extensor muscles; likewise, the posterior cord supplies extensors of the arm, forearm, and hand. The lateral cord (C5, C6, and C7) gives origin to the lateral pectoral nerve, the musculocutaneous nerve, and the lateral root of the median nerve. There is no sensory loss in the limb in this patient; injury to any of the other nerve elements listed here would be associated with specific dermatome losses.

So the answer to this question is C


Instruction:
Calculate the deficit for a 60 kg person,with Hb - 5 g/dl add 1000 mg for iron stores.
A. 1500
B. 2500
C. 3500
D. 4000

Response:
Iron deficit = Bodyweight(kg) x 2.3 x (15 - Hb) + 1000
      = 60 x 2.3 x (15 - 5) + 1000
      = 2380 (close to 2500)

So the answer to this question is B


Instruction:
Berkesonian bias is due to?
A. Presence of confounding factors in both cases & controls
B. Questioning the cases more thoroughly as compared to controls
C. Different rates of admission to hospital due to different diseases
D. Better recall by the cases as compared to controls

Response:
ANSWER: (C) Different rates of admission to hospital due to different diseasesREF: Park 20th edition page 68, 69, style="color: rgba(0, 0, 0, 1); font-family: Times New Roman, Times, serif">, 18. pdf, http://carcin.oxford jour nals. org/content/19/12/2063. full, pdfBias is any systematic error in the determination of the association between the exposure and disease.A statistic is biased if it is calculated in such a way that it is systematically different from the population parameter of interest. The following lists some types of, or aspects of, bias which should not be considered mutually exclusive:Confounding bias: In statistics, a confounding variable (also confounding factor, lurking variable, a confound, or confounder) is an extraneous variable in a statistical model that correlates (positively or negatively) with both the dependent variable and the independent variable. This bias is removed by matching in case control studies.CONFOUNDINGNON CONFOUNDINGCigarette Smoking------ Pre Cancerous Lung Cells ------ Lung CancerII. Selection bias: The error introduced when the study population does not represent the targetpopulation. Examples of selection bias areSampling bias is systematic error due to a non-random sample of a population. It is mostly classified as a subtype of selection bias, sometimes specifically termed sample selection bias, but some classify it as a separate type of bias.Healthcare access bias: when the patients admitted to an institution do not represent the cases originated in the communityBerkesonian bias: first described by Berkson in 1946 for case-control studies. It is produced when the probability of hospitalisation of cases and controls differ.Neyman bias (synonyms: incidence-prevalence biast selective survival bias): This type of bias can occur in cross-selectional and case-control studies when asymptomatic, mild, clinically resolved, or fatal cases are inadvertently excluded from the case group because the selected cases are examined some time after the disease process has already begun (i.e, looking at prevalent versus incident cases).Healthy worker effect:Volunteer bias is another type of selection bias. It is due to the fact that those individuals who take part in epidemiologic studies tend to be systematically different from those who do not.7. Loss to follow-up bias is a form of selection bias that may occur when there are significant losses to follow-up in ongoing cohort and experimental studies. If the final sample is systematically different from the original or target population with regard to exposure frequency or outcome susceptibility, then spurious associations can result.III.Information/Observation bias/ Measurement Bias: occurs when informationis collecteddifferently between two groups, leading to an error in the conclusion of the association. Wheninformation is incorrect, there is misclassification. Examples of Information Bias areMemory/Recall bias: When cases and control are asked questions about their past history, it may be more likely for the cases to recall the existence of certain event or factors than the control who are healthy persons. Recall bias is a type of measurement bias that is very common in case-control studies. It can be reduced using blindingInterviewer bias: Interviewer knows the status of the subjects before the interview process. Interviewer may probe differently about exposures in the past if he or she knows the subjects as cases. Interviewer bias is a type of measurement bias that can arise in all types of epidemiologic studies. It can be reduced using blindingLead time bias is the bias that occurs when two tests for a disease are compared, and one test (the new, experimental one) diagnoses the disease earlier, but there is no effect on the outcome of the disease--it may appear that the test prolonged survival, when in fact it only resulted in earlier diagnosis when compared to traditional methods.Hawthorne effect: described in the 1920s in the Hawthorne plant of the Western Electric Company (Chicago, IL). It is an increase in productivity--or other outcome under study-- in participants who are aware of being observedSurveillance bias/Detection bias: The group with the known exposure or outcome may be followed more closely or longer than the comparison group.Misclassification bias: When subjects are erroneously classified with respect to either exposure or disease status.Differential misclassification occurs when the level of misclassification differs between the two groups.Random or Non-differential misclassification occurs when the level of misclassification does not differ between the two groups.

So the answer to this question is C


Instruction:
A patient presents with upper abdominal pain and vomiting. His pulse is 110/min and BP is 70/40 mm Hg. A diagnosis, of pancreatitis is made. What is the next appropriate management?
A. Immediate IV fluids
B. Removal of necrosed pancreatic tissues
C. Start enteral nutrition first
D. IV antibiotics

Response:
Ans. (a) Immediate IV fluidsRef: Sabiston Textbook of surgery, 20th ed; pg. 1525-1530.* Key step in management of Acute Pancreatitis is - IV fluids.

So the answer to this question is A


Instruction:
According to WHO technical series published, CPITN probe is also known as:
A. 611
B. 617
C. 621
D. 625

Response:
None

So the answer to this question is C


Instruction:
Which of the following statements is true of primary grade IV-V vesicoureteric reflux in young children?
A. Renal scarring usually begins in the midpolar regions
B. Postnatal scarring may occur even in the absence of urinary tract infections
C. Long-term outcome is comparable in patients treated with either antibiotic prophylaxis or surgery
D. Oral amoxicillin is the choice antibiotic for prophylaxis

Response:
Ans. is 'b' i.e. Postnatal scarring may occur even in the absence of urinary tract infections Also see Nov. 98 (Q. no. 117), June 2004 (Q. no. Ill) Nov 2004 (Q. No. 122)Although UTI is the most important cause for scarring and nephropathy in pts. of VUR, scarring can occur even in the absence of UTI due to pressure-effect of reflux on the renal tissue.Other optionsOption a : Renal scarring usually begins in the polar region of the kidneys.Option d: Trimethoprim - sulfamethoxazole (Septra, Bactrim) is the DOC for prophylaxis.* Prophylactic AntibioticAmoxycillin or ampicillin is DOC for prophylaxis for children upto 6 wks of age After 6 wks, trimethoprim-sulfamethoxazole is the prophylactic DOC [this is because after 6 wks, the biliary system of the newborn becomes mature enough to handle trimethoprim-sulfamethoxazole. Nitrofurantoin (Macrodantin) is another acceptable option for prophylaxis.Medications are usually given as suspensions once daily at one-half the standard therepeutic dose. Night-time dosing in toilet trained children is most effective because it precedes the longest period of urinary retention, when infection is most likely to develop.Options in children who develop allergy to above drugs are - nalidixic acid, trimethoprim alone and cephalosporium.Option c : Grade IV-V primary VUR is unlikely to resolve spontaneously particularly if detected in children more than 1 yr of age. A medical management in them is usually ineffective.

So the answer to this question is B


Instruction:
Which of the following pathways does not occur in mitochondria
A. Beta oxidation
B. Urea cycle
C. Fatty acid synthesis
D. Heme synthesis

Response:
Fatty acid synthesis takes place in cytoplasm.
Beta Oxidation occurs in mitochondria.
Heme synthesis and Urea cycle occur both in cytoplasm and mitochondria.

So the answer to this question is C


Instruction:
All are true about Rinne's test except -
A. Positive in normal ear
B. Positive in sensorineural hearing loss
C. Minimum 15-20 dB air bone gap is required in conductive deafness
D. Bone conduction is better in sensorineural hearing loss

Response:
Ans. is 'd' i.e., Bone conduction is better in sensorineural hearing lossRinne's Testo The Rinne test is a tuning fork test that compares hearing by air conduction and bone conduction. The Rinne test is based on the idea that hearing mechanism is normally more efficient by air conduction (A C) than it is by bone conduction (BC), Le., AC > BC in normal persons. For this reason, a tuning fork will sound louder and longer by air conduction than by bone conduction. However, this air conduction advantage is lost when there is a conductive hearing loss in which case the tuning fork sounds louder by bone conduction than by air-conduction.Methodo Administering the Rinne test involves asking the patient to indicate whether a vibrating tuning fork sounds louder when its base is held against mastoid process (bone conduction) or when its prongs are held near the pinna, facing the opening of ear canal (air conduction). After striking the fork, the clinician alternates it between these two positions so that the patient can make a judgement about which one is louder.Interpretation of Rinne's testo The outcome of the Rinne test is traditionally called "positive" if the fork is louder by air conduction and this finding implies that the ear is normal or has sensorineural hearing loss. The results are called "negative" if the bone-conduction is louder than air-conduction, which is interpreted as revealing the presence of conductive deafness i.e., lesions of either external ear, tympanic membrane, middle ear or ear ossicles., o In conductive deafness Rinne's test will be negative if the conductive hearing loss is greater than 15-20 dB (minimum air bone gap 15-20 dB). That means, at least 15-20 dB of conductive hear loss is required to show bone conduction better than air conduction, Le., Negative Rinne's test. A negative Rinne test for 256, 512 and 1024 Hz indicates a minimum air-bone gap (ABG) of 15,30,45 dB respectively. Therefore, AB gap can be made if tuning forks of256,512 and 1024 Hz are used: -A Rinne test equal or negative for 256 Hz but positive for 512 Hz indicates air-bone gap of20-30 dB.A Rinne test negative for 256 and 512 Hz but positive for 1024 Hz indicates air-bone gap of30-45 dB.A Rinne negative for all the three tuning forks of256, 512, and 1024 Hz, indicates air-bone gap of 45- 60 dB.Rationale of positive testo Positive test (AC > BC) is seen in :-Normal personIt has already been explained that air conduction is better than bone conduction, thus Rinne test is positive in normal person.Sensori-neural hearingIn sensorineural hearing loss, the defect is in cochlea or VIII nerve or its central connection. There is no defect in conductive apparatus of ear (air conduction) or in mastoid bone (bone conduction). As air conduction is better than bone conduction, Rinne test will be positive (because both air conductive apparatus and bone conductive apparatus are normal and the pathology is distal to them).o Negative test (BC > AC) is seen in: -Conductive deafnessAs the conductive apparatus is defective, bone conduction becomes better than air conduction (BC > AC).Severe unilateral sensorineural hearing lossHere the Rinne test is false negative (not true negative) as it is interpreted by the patient that BC > AC, but actually it is not. In reality this response is from the opposite ear because of transcranial transmission of sound during bone conduction testing. This can be prevented by masking the non-test ear with Barany's box while testing the bone conduction. Weber test will help for such situation.

So the answer to this question is D


Instruction:
Cause of transudate pleural effusionn is
A. Nephrotic syndrome
B. Tuberculosis
C. Pneumonia
D. Pulmonary infarction

Response:
Causes of transudative effusion includes Nephrotic syndrome,congestive hea failure,cirrhosis,pulmonary embolization, myxedema,etc.Causes of exudative effusion includes neoplasms;infections like pneumonia,TB,collagen vascular diseases,Meigs syndrome,etc. Most common cause of exudative pleural effusion is bacterial pneumonia. Reference:Davidson's Medicine-22nd edition,page no:661.

So the answer to this question is A


Instruction:
A multipara with previous LSCS comes at 38 weeks pregnancy in shock. Differential diagnosis includes:
A. Placenta previa
B. Abruptio placenta
C. Rupture uterus
D. All of the above

Response:
Shock in Obstetrics:
Causes of shock during pregnancy

So the answer to this question is D


Instruction:
A subcentre in a hilly area caters to a population of -
A. 1000
B. 2000
C. 3000
D. 5000

Response:
The subcentre is the peripheral outpost of the existing health delivery system in rural areas .They are being established on the basis of one subcenter for every 5000 population in general and one for every 3000 population in hilly,tribal and backward areas (refer pgno:903 park 23rd edition)

So the answer to this question is C


Instruction:
False about facial muscles:
A. Dilates and constrict facial orifices
B. Supplied by facial nerve
C. Develops from 3rd pharyngeal arch
D. They develop from mesoderm

Response:
The muscles of the face are embedded in the superficial fascia, and most arise from the bones of the skull and are inseed into the skin.The orifices of the face, namely, the orbit, nose, and mouth, are guarded by the eyelids, nostrils, and lips, respectively.It is the function of the facial muscles to serve as sphincters or dilators of these structures.A secondary function of the facial muscles is to modify the expression of the face.All the muscles of the face are developed from the second pharyngeal arch and are supplied by the facial nerve.

So the answer to this question is C


Instruction:
The following are indications for performing thoractomy after blunt injury of the chest, except
A. 1500 ml drainage after placing an intercostal tube
B. Continuous bleeding through intercostal tube more than 200 ml\/hour for three or more hours
C. Cardiac tamponade
D. Rib fracture

Response:
Indications for thoracotomy include haemothorax more than 1500 ml after placing intercostal tube(ICT)or hourly collection of 200-300 ml or ICT placed showing persistent drainage of blood.haemopericardium leading to cardiac tamponade is an another indication.Other indications include diaphragmatic injury,bronchus & major vessel injuries,oesophageal & thoracic duct injuries.

So the answer to this question is D


Instruction:
Which of the following drugs is contra-indicated in acute intermittent porphyria?
A. Thiopentone
B. Midazolam
C. Propofol
D. Etomidate

Response:
Barbiturates are contra-indicated in acute intermittent porphyria. Ref: KD Tripathi 8th ed.

So the answer to this question is A


Instruction:
Drug of choice for Prophylaxis of Bronchialitis in a child with heart disease
A. Ribavirin
B. Doxycycline
C. Palivizumab
D. Penicillin V

Response:
Treatment for Bronchiolitis in a child with co-morbidities is inhaled Ribavirin.
Drug of choice for prophylaxis of Bronchialitis in a child with co-morbidities is Palivizumab.

So the answer to this question is C


Instruction:
The enzyme not present in the skeletal muscles :
A. Creatinine phosphokinase
B. Hexokinase
C. Phosphofructokinase
D. Glucose 6 phosphatase

Response:
D i.e. Glucose -6 - phosphataseThe role of glycogen in liver is to provide free glucose for expo to maintain the blood glucose level; whereas in muscles the role of glycogen is to provide a source of glucose-6-PO4 for glycolysis (to supply ATP for muscle contraction).This difference of fate of glycogen in liver & muscle is d/t absence of glucose -6- phosphatase enzyme in muscleQ. Because of absence of this enzyme, which catalyzes the final step of glycogenolysis, glycogen in muscles can only be conveed upto glucose-6- phosphate. Hence muscle glycogenolysis does not contribute to blood glucose directlyQ. Instead, in muscle, this glucose -6-PO4 can not cross plasma membrane and enters glycolysis producing energy (3ATP) needed for muscle contraction. The gain of an extra (1) ATP is d/t bypass of 1st step of glycolysis which requires 1 ATP (If glycolysis stas from glucose only 2 ATPs are produced). Lactic acid produced d/t glycolysis crosses the plasma membrane & reaches liver to form glucose.In liver glucose -6. phosphate is transpoed by glucose- 6-phosphate translocase into endoplasmic reticulum, where it is conveed to glucose by glucose- 6- phosphatase (this enzyme is also used in last step of gluconeogenesis). The glucose moves out of ER to cytosol and then released by hepatocytes into blood to maintin blood glucose levels untilgluconeogenesis stas.

So the answer to this question is D


Instruction:
Type I hypersensitivity includes all of the following except:
A. Autoimmune hemolytic anemia
B. Anaphylaxis
C. Extrinsic asthma
D. Hay fever

Response:
Ans. is 'a' i.e., Autoimmune hemolytic anemia (Ref: Ananthanarayan, 9th/e, p. 162 and 8th/e, p. 162)* Autoimmune hemolytic anemia is type II HSN.* In type II Hypersensitivity reaction (Ab mediated) various mechanisms involves up: (1) Opsonization, complement and Fc receptor mediated phagocytosis (C3b acts as opsonin). (2) Complement and Fc receptor mediated inflammation (C5a acts as chemokine for neutrophil and monocyte. (3) Antibody mediated cellular dysfunction.

So the answer to this question is A


Instruction:
Dyskeratosis is characteristic feature of :
A. Darrier's ds
B. Pemphigus vulgaris
C. Psoriasis
D. All

Response:
A i.e. Darrier's ds

So the answer to this question is A


Instruction:
AFP surveillance registry indicator is ?
A. Number of AFP cases repoed
B. Number of wild polio-virus positive cases
C. Number of non-polio AFP < 5 years
D. Number of non-polio AFP < 15 years

Response:
Ans. is 'a' i.e., Number of AFP cases repoed The number of AFP cases repoed each year is used as an indicator of a country's ability to detect polio, even in countries where the disease no longer occurs. Polio surveillance It is the most impoant pa of whole polio eradication intiative. It has two components:? Acute flaccid paralysis (AFP) surveillance Acute flaccid paralysis is defined as acute onset (< 4 weeks) of flaccid paralysis (reduced tone) without other obvious cause in children WHO recommends the immediate repoing and investigation of every case of AFP in children less than 15 years.

So the answer to this question is A


Instruction:
Treatment of chronic dacrocystitis:
A. Dacrocystorhinostomy
B. Antibiotics
C. Probing/Syringing
D. All

Response:
A i.e. Dacrocystorhinostomy; B i.e. Antibiotics; C.ie Probing Congenital dacryocystitis is almost always chronic Q, while acquired may be acute or chronic (chronic being more common). It is treated by massage over lacrimal sac, lacrimal syringing with antibiotics, probing of nasolacriinal duct, antibiotic drops and surgery (dacryocustorhinostomy, dacryocystectomy, conjunctivocystorhinostomy) Q

So the answer to this question is D


Instruction:
Which is associated with shotgun alone -
A. Gun powder
B. Primer
C. Projectile
D. Wads

Response:
Wads are present only in the cartridge of a shot gun.

So the answer to this question is D


Instruction:
Transmission of cholera is through ?
A. Fecally contaminated food
B. Fecally contaminated water
C. Contaminated food by vomits of a case
D. All of the above

Response:
Man is the only reservoir. The immediate source of infection are the stools and vomits of cases and carrier. Infection is acquired through fecally contaminated water or food. Chlorination of water is effective against V cholerae. There are following tyes of carrier in cholera. i) Incubatory : Shed vibrios only in the brief incubation period of 1-5 days. ii) Convalescent : Shed vibrios for 2-3 weeks. iii) Healthy or contact carrier : Has had subclinical infection and shed vibrios for less than 10 days. iv) Chronic carriers : Can shed vibrios for months or years and may have persistent infection in gall bladder.

So the answer to this question is D


Instruction:
who gave the psychosocial stage of development
A. erikson
B. bleuler
C. freud
D. lorenz

Response:
ERIC ERIKSON ERIC ERIKSON gave the psychosocail stages of development There are 8 psychosocial stages A person has to succesfully pass one stage to move to the next stage If he gets arrested in one stage he may develop some disorders Ref. kaplon and sadock, synopsis of psychiatry, 11 th edition, pf no. 485

So the answer to this question is A


Instruction:
Best prognosis for reversibility is seen in :
A. lsthmo -- isthmic type
B. Isthmic -- ampullary type
C. Ampullary -- interstitial type
D. Ampullary -- timbrial type

Response:
Ans. is a i.e. isthmo -isthmic type Read the following lines "lt is impoant to select the site of tuba! ligation carefully which should ideally be done at the tubal isthmus. This is because in the event of the patient desiring a tubal recanalization procedure, the isthmoisthmic anastomosis carries the best chances of success". Site for performing tubal ligation = isthmusdeg. Postpaum laparoscopic ligation - done after 6 weeks postpaumdeg.

So the answer to this question is A


Instruction:
Plasma volume is best evaluated by using?
A. Evan blude
B. Inulin
C. Mannitol
D. Radoplabelled water

Response:
Plasma volume is measured using dyes like Evans blue that binds specifically to plasma proteins. (Ref: Textbook of physiology N Geetha 2 nd edition page no.62)

So the answer to this question is A


Instruction:
A female patient is diagnosed with primary herpes genitalis. Which of the following should be the treatment advice?
A. "Apply one applicator of Terconazole intravaginally at bedtime for 7 days."
B. "Apply one applicator of Sulconazole intravaginally at bedtime for 7 days."
C. Acyclovir 200mg ; five times a day for 7 days."
D. "Apply Acyclovir twice daily by massaging it gently into the lesions for 3 days."

Response:
A patient with primary herpes genitalis : Acyclovir 200mg ; 5 times a day for 7 days Or 400mg, 3 times a day for 7 days. Terconazole and Tioconazole are used to treat vulvovaginal candidiasis. Sulconazole nitrate is used to treat tinea versicolor.

So the answer to this question is C


Instruction:
A 29-year-old man who has had no major illnesses experiences acute onset of hemoptysis. On physical examination, he has a temperature of 37deg C, pulse of 83/min, respirations of 28/min, and blood pressure of 150/95 mm Hg. A chest radiograph shows bilateral fluffy infiltrates. A transbronchial lung biopsy on microscopic examination shows focal necrosis of alveolar walls associated with prominent intra-alveolar hemorrhage. Two days later, he has oliguria. The serum creatinine level is 2.9 mg/dL, and urea nitrogen is 31 mg/dL. Which of the following antibodies is most likely involved in the pathogenesis of his condition?
A. Anti-DNA topoisomerase I antibody
B. Anti-glomerular basement membrane antibody
C. Antimitochondrial antibody
D. Anti-neutrophil cytoplasmic antibody

Response:
Goodpasture syndrome leads to renal and pulmonary lesions produced by an antibody directed against an antigen common to the basement membrane in glomerulus and alveolus. This leads to a type II hypersensitivity reaction. The anti-DNA topoisomerase I antibody is a marker for scleroderma. Antimitochondrial antibody is associated with primary biliary cirrhosis. C-ANCA and P-ANCA are best known as markers for various forms of systemic vasculitis. ANA is used as a general screening test for various autoimmune conditions, typically collagen vascular diseases such as systemic lupus erythematosus.

So the answer to this question is B


Instruction:
In a case of chest pain with pericarditis and pericardial effussion, pain is referred by?
A. Phrenic nerve
B. Superficial cardiac plexus
C. Deep cardiac plexus
D. Vagus nerve

Response:
Phrenic nerveThe pericardium is composed of two layers.The tough external layer, the fibrous pericardium. The parietal layer of serous pericardium is attached to its internal surface. The visceral layer of serous pericardium is fused to the hea.The fibrous and parietal pericardia are supplied by the phrenic nerve (C3-05). They are sensitive to pain.The visceral pericardium or epicardium is supplied by autonomic nerves, not sensitive to pain.

So the answer to this question is A


Instruction:
All of the following are topically used sulfonamides EXCEPT:
A. Sulfacetamide
B. Sulfasalazine
C. Silver sulfadiazine
D. Mafenide

Response:
SULFONAMIDES: 1 . Sho acting ( 4-8 hr ): Sulfadiazine 2. Intermediate acting (8-12 hr ): Sulfamethm,azole 3. Long acting ( -7 days): Sulfadoxine, Sulfamethopyrazine 4. Special purpose sulfonamides: Sulfacetamide sod, Mafenide, Silver sulfadiazine, Sulfasalazine. Sulfasalazine (Salicylazosulfapyridine) It is a compound of 5-aminosalicylic acid (5-ASA) with sulfapyridine linked through an azo bond that has a specific therapeutic effect in inflammatory bowel diseases (IBDs) like ulcerative colitis and Crohn&;s disease.They are both locally and systemically acting agents. ESSENTIALS OF MEDICAL PHARMACOLOGY K.D.TRIPATHI SIXTH EDITION PAGE NO:661

So the answer to this question is B


Instruction:
Which of the following is an intermediate level disinfectant ?
A. 2% glutarldyhyde
B. Ethylene oxide
C. Hypochlorite
D. None

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

So the answer to this question is C


Instruction:
Eruption of Deciduous Mandibular Central incisor occur at
A. 8-9 months
B. 6-7 months
C. 19-20 months
D. 15-16 months

Response:
None

So the answer to this question is B


Instruction:
Facial recess is bounded by:
A. Medially by the vertical part of VII nerve
B. Laterally by the chorda tympani
C. Above by fossa incudis
D. All of the above

Response:
(d) All of the above(Ref. Cummings, 6th ed., 509; 2192)Facial recess is an area on the posterior wall of middle ear. The facial recess is limited superiorly by the fossa incudis, laterally by chorda tympani entry and medially by the descending/vertical facial nerve segment.The facial recess is the site where opening is made on the posterior wall to access the middle ear cavity through the mastoid in "INTACT CANAL WALL" ear surgeries. This is known as the posterior tympanotomy approach.

So the answer to this question is D


Instruction:
According to the ICD-10 revision, for establishing a diagnosis of mania, the symptoms should persist for at least:
A. 1 week
B. 2 weeks
C. 3 weeks
D. 4 weeks

Response:
Ans. a. 1 week (Ref: Niraj Alhuja 6/e p75- 76)According to the ICD-10 revision, for establishing a diagnosis of mania, the symptoms should persist for at least 1 week.Diagnostic Criteria of Manic EpisodeAbnormality and persistently elevated, irritable or expensive mood for >1 weekQ or need hospitalization>=3 of these hyperactivities:Decreased need of sleepsIncrease in goal orientated activityQ (social, spiritual, sexual, religious)Excessive involvement in pleasurable activitiesQ (unrestrained buying sprees, foolish business investments, donations or sexual indiscretion) that have a higher potential for painful consequencesMore talkative or pressure to keep talkingQInflated self esteem or grandiosityQDistractibilityQFlight of ideas or racing thoughtsQ

So the answer to this question is A


Instruction:
Schaumann bodies are seen in:
A. Sarcoidosis
B. Chronic bronchitis
C. Asthma
D. Syphilis

Response:
Ans. (a) Sarcoidosis(Ref: Robbins 9th/pg 693-694)Histology in Sarcoidosis:Well-formed non-caseating granulomas Q composed of an aggregate of tightly clustered epithelioid cells Q, with Langhans or foreign body-type giant cells; Central necrosis is unusual. QSchaumann bodies: laminated concretions composed of calcium and proteins QAsteroid bodies: Stellate inclusions enclosed within giant cells Q

So the answer to this question is A


Instruction:
Trichotillomania?
A. Irresistable desire to set fire
B. Irresistable desire to steal things
C. Compulsive hair pulling
D. Pathological gambling

Response:
Ans. is 'c' i.e., Compulsive hair pulling

So the answer to this question is C


Instruction:
Which of the following drugs is not a part of the 'Triple Therapy' immunosuppression for post-renal transplant patients?
A. Cyclosporine
B. Azathioprine
C. FK 506
D. Prednisolone

Response:
nitial therapy is the treatment given to all recipients (except where the donor is an identical twin) for 0–3 months after transplantation.
Initial therapy is usually 'triple therapy', in which a calcineurin inhibitor (traditionally ciclosporin) is used as the 'primary agent'in combination with a corticosteroid (prednisolone) and azathioprine.

So the answer to this question is C


Instruction:
Which of the following markers is specific for gastro-intestinal stomal tumor (GIST)
A. CD117
B. CD34
C. CD23
D. S-100

Response:
most useful diagnostic marker of GIST is is c-KIT or CD117 REF :SYSTEMIC AND GENERAL PATHOLOGY RAMDAS NAYAK 2ND EDITION PAGE NO:390 Refer robbins 9/e

So the answer to this question is A


Instruction:
Preferred fluid in a poly-traumatic patient with shock?
A. Ringer lactate
B. Normal saline
C. Dextran
D. Dextrose-normal saline

Response:
Ans. (b) Normal salineRef.-.Harrison 18th ed. / 270Normal saline is the preferred fluid for volume expansion in a crashing patient with un-recordable BP. The main point to be noted is that if ringer lactate is infused in these patients, then lactate in the solution will never reach the liver where it is normally converted into bicarbonate. Hence the accumulated lactate will get broken into lactic acid that will worsen the status of patient of poly-trauma who is already acidotic. Dextran would not be indicated as it exhibits anti-thrombotic tendency.

So the answer to this question is B


Instruction:
Which of the following does not help in the prevention of reflux esophagitis?
A. Long intraabdominal esophagus
B. Increased intraabdominal pressure
C. Right crus of diaphragm
D. Increased intrathoracic pressure

Response:
The principal barrier to reflux is the lower esophageal sphincter. LES comprises the lower 4 cm of esophagus, where resting pressure within the lumen normally exceeds the intragastric pressure. This high intraluminal pressure at LES prevents the GE reflux. The LES is a physiological entity rather than any anatomical structure. Factors contribute to the high intraluminal pressure of LES: Intrinsic musculature of distal esophagus: these muscles differ from those in the other area of the esophagus in that they are in a state of tonic contraction. They normally relax with initiation of the a swallow and then return to a state of tonic contraction. Sling fibres of cardia Crura of the diaphragm Intra Abdominal pressure: the intra abdominal pressure compresses the abdominal segment of the esophagus and is probably the most impoant factor in preventing the reflux. The gastroesophageal angle: the oblique entry of the esophagus has a valvular effect, in that it closes the esophagus when gastric fundus distends. Folds of the mucosa at the cardia : helps in closing the LES when intragastric pressure arises. Intrathoracic pressure has no role in preventing reflux as it is too low ranging from -5 to + 5mm of Hg Ref: Sabiston 18/e, Page 1108 ; Essential of Surgery By Sunil Chumber 1/e, Page 7 ; Textbook of Surgery By S. Das 3/e, Page 790

So the answer to this question is D


Instruction:
'Obstruction of Inferior vena cava' presents -
A. Paraumbilical dilatation
B. Thoracoepigastric dilatation
C. Oesophageal varices
D. Haemorrhoids

Response:
Obstruction of inferior vena cava presents as thoracoepigastric dilatation because the vessels involved are the superficial veins, namely, superficial epigastric, lateral thoracic, thoracoepigastric.

So the answer to this question is B


Instruction:
The term "Guyon&;s canal" is in relation to which of the following?
A. Anterior interosseus nerve
B. Ulnar nerve
C. Median nerve
D. Radial aery

Response:
Guyon's canal or ulnar tunnel is an osseofibrous tunnel formed by the pisohamate ligament bridging the concavity formed by the pisiform bone and hook of the hamate. The ulnar a nerve passes through this tunnel in the wrist and its compression here lead to ulnar tunnel syndrome characterised by hypoesthesia of medial 1 and 1/2 fingers and weakness of intrinsic muscles of the hand. Ref: Textbook of anatomy upper limb and thorax, Vishram Singh, 2nd edition, page no.150, 180

So the answer to this question is B


Instruction:
B cell prolymphocytic leukemia patients differ from those with B cell chronic lymphocytic leukemia in -
A. Presenting at a younger age
B. Having a lower total leucocyte count
C. Having prominent lymphadenopathy
D. Having a shoer survival

Response:
cell prolymphocytic leukemia (B-PLL) is a very rare B cell neoplasm comprised of so-called prolymphocytes, typically with involvement of the peripheral blood, bone marrow, and spleen. The name "prolymphocyte" is actually a misnomer, as the tumor cells in this disease are mature activated B cells The 5-year survival rate for people age 20 and older with CLL is 84%. It is impoant to remember that statistics on the survival rates for people with leukemia are an estimate. B-cellchronic lymphocytic leukaemia (CLL) develops from a type of white blood cell called B cells. It progresses slowly, usually affecting older adults. Survival rate for chronic lymphocytic leukemia CLL has a higher survival rate than many other cancers. The five-year survival rate is around 83 percent. This means that 83 percent of people with the condition are alive five years after diagnosis Ref Davidson 23rd edition pg 926-927

So the answer to this question is D


Instruction:
Telefona
A. Pulling of hair
B. Beating on soles
C. Beating on both ears
D. Beating on head

Response:
C i.e. Beating of both ears Telefona is simultaneous beating of both ears by both palms of hands, which may lead to torn external ear, rupture of tympanic membrane causing pain, bleeding & hearing loss.

So the answer to this question is C


Instruction:
Narrowest pa of uterine tube is: March 2013
A. Infundibulum
B. Ampulla
C. Isthmus
D. Interstitial

Response:
Ans. D i.e. Interstitial Fallopian tube Lined by: Ciliated columnar epithelium Length: 10 cms

So the answer to this question is D


Instruction:
How is the depth of curing and shade of the material related?
A. Lighter shade requires more curing
B. Darker shade requires less curing
C. Lighter shade requires less curing
D. Both 1 and 2

Response:
Optical modifiers affect light transmission through a composite. Thus, darker shades and greater opacities have a decreased depth of light-curing ability and require either an increased exposure time or a thinner layer when cured. Studies have shown that for optimal polymerization, resins with darker shades and opacifiers should be placed in thinner layers.
Phillip’s 12th edition page 291

So the answer to this question is C


Instruction:
Hexose sugar is not present in:
A. Ribose
B. Glucose
C. Fructose
D. Galactose

Response:
Hexose sugar is present in:Glucose, "blood sugar", the immediate source of energy for cellular respirationGalactose, a sugar in milk (and yogu), andFructose, a sugar found in honeyRibose is a pentose sugar

So the answer to this question is A


Instruction:
Triangular space between clitoris and hymen -
A. Fourchette
B. Fossa navicularis
C. Vestibule
D. Labia minora

Response:
Ref: synopsis of Forensic Medicine & Toxicology (K.S.Narayan Reddy)28th edition ,page 203 In a virgin female, the labia minora are two thin folds of skin just within the labia majora. The lower poions of labia mainora fuse in midline and form a fold fourchette. The depression between fourchette and vaginal orifice is called fosa navicularis. The clitoris is small and the triangular narrow space clitoris and hymen is called vestibule.

So the answer to this question is C


Instruction:
A 24-year-old male repos that during a game of hoops (basketball), he tripped while driving the ball to the basket, and fell on his outstretched right hand with the palm down. Three days later, patient presented to the OPD with right wrist pain. On examination, the wrist was slightly swollen and tender but without deformity. The doctor instructed the patient to extend the right thumb, thereby accentuating the anatomical "snuffbox," which is extremely tender to deep palpation. The doctor advised to get a x ray of the wrist done. What is the most likely injured pa?
A. D
B. A
C. B
D. C

Response:
A- Hamate B- Pisiform C- Lunate D- scaphoid This young man tripped while playing basketball and stretched out his right hand to protect himself. His hand, with the palm down and probably deted to the side of the radius, took the brunt of the fall, resulting in significant impact force to the wrist. This resulted in pain and swelling of the wrist, especially on the radial side, with point tenderness deep in the anatomical snuffbox. This is the common mechanism for a fracture of the scaphoid carpal bone, the most commonly fractured carpal bone. Point tenderness over a bone or bony process is a hallmark of a fracture at that site. Radiologic confirmation of a fracture is impoant. The scaphoid bone has a unique blood supply, and proper reduction and alignment of the segments is necessary to decrease the risk of avascular necrosis.

So the answer to this question is A


Instruction:
Maximum cholesterol is seen in
A. VLDL
B. LDL
C. HDL
D. Chylomicrons

Response:
Ref: Textbook of Medical Biochemistry Dr (Brig) MN Chatterjea Rana Shinde, 8th Edition, page no: 446

So the answer to this question is B


Instruction:
Use of dimethyl-P-toluidine is indicated for
A. Thermal polymerization of acrylic
B. Chemical polymerization
C. Retarding the polymerization reaction
D. To inhibit the action of benzoyl peroxide

Response:
None

So the answer to this question is B


Instruction:
Which of the following drugs is most frequently used in treating refractory periodontitis, including localized juvenile periodontitis?
A. Metronidazole and Amoxicillin
B. Tetracycline
C. Ciprofloxacin
D. Penicillin

Response:
None

So the answer to this question is A


Instruction:
Apolipoprotein B48
A. RNA alternate splicing
B. RNA editing
C. DNA editing
D. RNA inference

Response:
REF : LIPINCOTT 6TH ED

So the answer to this question is B


Instruction:
A 25-year-old man comes to the physician because of headache, dizziness, and bilateral leg claudication, Physical examination shows hypertension in the upper limbs and hypotension in the lower limbs.

Which of the following additional findings is mostly likely in this patient?
A. Aortic valvular stenosis
B. Notching of inferior margins of ribs
C. Patent ductus arteriosus
D. Vasculitis involving the aortic arch

Response:
None

So the answer to this question is B


Instruction:
Best investigation when there is clinical suspicio of pulmonary embolism in a patient is -
A. D - Dimer assay
B. Multi detector CT angiography
C. Doppler ultrasound
D. Catheter angiography

Response:
None

So the answer to this question is B


Instruction:
A 28 Years of old lady presented with wrist pain. X-ray wrist us showing lytic eccentric lesion in lower end of radius with soap bubble appaerance, what is the next plan for managemnt
A. Bone curettage and bone grafting
B. Extended curettage with phenol
C. Biopsy of lesion
D. Extended curettage with phenol and bone grafting

Response:
It is a case of giant cell tumor Actually next investingation should be MRI followed by biopsy In bone tumors :x ray ------MRI-----biopsy------treatment ref : maheswari 9th ed

So the answer to this question is C


Instruction:
Aortic opening in diaphragm is found at the level of:
A. T12
B. T10
C. T8
D. LI

Response:
Ref: Grays Anatomy, 40th edExplanation:A number of structures pass between the thorax and abdomen via apertures in the diaphragm.There are three large openings, for the aorta, oesophagus and inferior vena cava, and a number of smaller ones.The aortic apertureIt is the lowest and most posterior of the large openings, and is found at the level of the lower border of the twelfth thoracic vertebra and the thoracolumbar intervertebral disc, slightly to the left of the midline.The aortic opening transmits the aorta, thoracic duct, lymphatic trunks from the lower posterior thoracic wall and. sometimes, the azygos and hemiazygos veins.The oesophageal apertureIt is located at the level of the tenth thoracic vertebra, above, in front and a little to the left of, the aortic opening.It transmits the oesophagus, gastric nerves, oesophageal branches of the left gastric vessels and some lymphatic vessels.The vena caval apertureThe highest of the three large openings, lies at about the level of the disc between the eighth and ninth thoracic vertebrae.It is traversed by the inferior vena cava, which adheres to the margin of the opening, and by some branches of the right phrenic nerve.Also note:There are two lesser apertures in each crus: one transmits the greater, and the other the lesser, splanchnic nerve.The ganglionated sympathetic trunks usually enter the abdominal cavity behind the diaphragm, deep to the medial arcuate ligament.Openings for minute veins frequently occur in the central tendon.* Aortic, Thoracic duct, Azygos Vein opening -112* Oesophageal opening -T10* Inferior venacava opening-T8

So the answer to this question is A


Instruction:
Which one of the following drugs is not contraindicated in bronchial asthma?
A. Morphine
B. Propanolol
C. Stropine
D. Nitroglycerine

Response:
None

So the answer to this question is D


Instruction:
Diapedesis is :
A. Immigration of leucocytes through the basement membrane
B. Immigration of the leucocytes through the vessel wall to the site of inflammation
C. Aggregation of platelets at the site of bleeding
D. Auto digestion of the cells

Response:
. Immigration of the leucocytes through the vessel wall to the site of inflammation

So the answer to this question is B


Instruction:
Contraindications to major hepatic resection for metastatic disease include all of the following except-
A. Total hepatic involvement
B. Advanced cirrhosis
C. Extrahepatic tumor involvement
D. Jaundice from extrinsic ductal obstruction

Response:
None

So the answer to this question is D


Instruction:
A 30 year old female epileptic on phenytoin, developed weakness and fatigue. Blood examination revealed Hb = 4.6 gm. MCV = 102 fl and MCH = 40 pg/dl. What is the most probable diagnosis ?
A. Hea failure
B. Iron deficiency anaemia
C. Phenytoin induced agranulocytosis
D. Megalobalstic anaemia

Response:
Ans. is 'd' i.e., Megalablostic anaemia Raised MCV (normal 86-98) and MCH (normal 28-33) with low hemoglobin (normal 12-16) suggest the diagnosis of megaloblastic anemia. Remember Amongst antiepileptics, phenytoin, phenobarbitone and primidone can cause megaloblastic anemia.

So the answer to this question is D


Instruction:
Concentration of MgSO4 in the treatment of eclampsia in meq/L:
A. 7 − 10.
B. 10 − 15.
C. 2 − 4.
D. 4 − 7.

Response:
None

So the answer to this question is D


Instruction:
A freshly prepared oral rehydration solution should not be usved after -
A. 4 hrs
B. 6 hrs
C. 12 hrs
D. 24 hrs

Response:
ORS Aim of oral rehydration therapy is to prevent dehydration and reduce moality. Oral fluid therapy is based on the observation that glucose given orally enhances the intestinal absorption of salt and water and is capable of correcting electrolyte and water deficit. At first the composition of oral rehydration salt recommended by WHO was sodium bicarbonate based. Inclusion of trisodium citrate in place of sodium bicarbonate made the product more stable and it resulted in less stool output especially in high output diarrhoea, probably because of direct effect of trisodium citrate in increasing intestinal absorption of sodium and water. More recently an improved ORS formulation has been developed which is safe and effective as the original in preventing and treating diarrhoeal dehydration. It is focussed on reducing the osmolarity of ORS solution to avoid adverse effects of hypeonicity on net fluid absorption by reducing the concentration of glucose and sodium chloride in the solution. A freshly prepared ORS should be used before 24hrs. Decreasing the sodium concentration of ORS solution to 75 mOsm/l improved the efficacy of ORS regimen for children with acute non cholera diarrhoea. Reduced osmolarity ORS:-(g/L) Sodium chloride- 2.6 Glucose,anhydrous- 13.5 Potassium chloride- 1.5 Trisodium citrate,dihydrate-2.9 Total =20.5 g/L Reduced osmolarity ORS:-(mmol/L) Sodium-75 Chloride-65 Glucose,anhydrous-75 Potassium-20 Citrate-10 Total = 245 mmol/L {Reference: park&;s textbook of preventive and social medicine, 23rd edition, pg no.224}

So the answer to this question is D


Instruction:
Aerial supply of tonsil is mainly by:
A. Maxillary aery
B. Tonsillar branch of facial aery
C. Middle meningeal aery
D. Internal carotid aery

Response:
Ans. B Facial aery The aerial supply to the tonsil derives mainly from the tonsillar branch of the facial aery and the descending palatine aery; veins drain into the pharyngeal plexus and the lymphatic drainage from the tonsil is into the upper deep cervical lymph nodes, which may enlarge during tonsillar infection. The tonsil is supplied by five aeries a. Tonsillar branch of facial- This is the main aery b. Ascending pharyngeal aery from external carotid c. Ascending palatine - a branch of facial aery d. Dorsal linguae branches of lingual aery e. Descending palatine branch of the maxillary aery.

So the answer to this question is B


Instruction:
Fan shaped muscle of tongue, which forms the main bulk:
A. Hyoglossus.
B. Genioglossus.
C. Verticalis.
D. Palatoglossus.

Response:
None

So the answer to this question is B


Instruction:
Serum amylase level is increased in all of the following except: March 2005
A. Intestinal obstruction
B. Perforated ulcer
C. Acute pancreatitis
D. Acute appendicitis

Response:
Ans. D: Acute appendicitis Increased amylase levels may indicate: - Acute pancreatitis - Cancer of the pancreas, ovaries, or lungs - Cholecystitis - Infection of the salivary glands (such as mumps) or an obstruction - Intestinal obstruction - Macroamylasemia Pancreatic or bile duct obstruction - Perforated ulcer - Tubal pregnancy (may be ruptured) - Viral gastroenteritis Decreased amylase levels may indicate: - Damage to the pancreas - Kidney disease - Pancreatic cancer - Toxemia of pregnancy Drugs that can increase amylase measurements include: - Asparaginase - Aspirin Bih control pills - Cholinergic medications - Coicosteroids - Indomethacin - Loop and thiazide diuretics - Methyldopa - Opiates (codeine, morphine) - Pentazocine

So the answer to this question is D


Instruction:
Tubercles of Montgomery are present in:
A. Lung
B. Duodenum
C. Liver
D. Breast

Response:
Montgomery’s glands or Montgomery Tubercles are the large sebaceous and rudimentary milk glands present in the areola surrounding the nipple of the female breast.

So the answer to this question is D


Instruction:
A 65-year-old man undergoes a technically difficult abdominal-perineal resection for a rectal cancer during which he receives 3 units of packed red blood cells. Four hours later, in the intensive care unit (ICU), he is bleeding heavily from his perineal wound. Emergency coagulation studies reveal normal prothrombin, paial thromboplastin, and bleeding times. The fibrin degradation products are not elevated, but the serum fibrinogen content is depressed and the platelet count is 70,000/mL. Which of the following is the most likely cause of his bleeding?
A. Delayed blood transfusion reaction
B. Autoimmune fibrinolysis
C. A bleeding blood vessel in the surgical field
D. Factor VIII deficiency

Response:
Whenever significant bleeding is noted in the early postoperative period, the presumption should always be that it is due to an error in surgical control of blood vessels in the operative field. Hematologic disorders that are not apparent during the long operation are most unlikely to surface as problems postoperatively. Blood transfusion reactions can cause diffuse loss of clot integrity; the sudden appearance of diffuse bleeding during an operation may be the only evidence of an intraoperative transfusion reaction. In the postoperative period, transfusion reactions usually present as unexplained fever, apprehension, and headache--all symptoms difficult to interpret in the early postoperative period. Factor VIII deficiency (hemophilia) would almost ceainly be known by history in a 65-year-old man, but, if not, intraoperative bleeding would have been a problem earlier in this long operation. Moreover, factor VIII deficiency causes prolongation of the paial thromboplastin time (PTT), which is normal in this patient. Severely hypothermic patients will not be able to form clots effectively, but clot dissolution does not occur. Care should be taken to prevent the development of hypothermia during long operations through the use of warmed intravenous fluid, gas humidifiers, and insulated skin barriers.

So the answer to this question is C


Instruction:
Basis of division of anatomical segments of liver are all, EXCEPT:
A. Hepatic aery
B. Hepatic vein
C. Poal vein
D. Bile duct

Response:
The proper hepatic aery, poal vein, and common hepatic duct enter/exit the liver at the poa hepatis as the 'poal triad'. The central branch first divides into two larger branches, functionally subdividing the liver into left pas and right pas. The poal triad continue to ramify within the liver, forming a total of eight segments that are more or less functionally independent of one another. Hepatic segments grouped by pas and divisions: Left pa of the liver 1. Posterior pa, Caudate lobe Segment (S1) 2. Left lateral division 1. Left posterolateral segment (S2) 2. Left anterolateral segment (S3) 3. Left medial division Left medial segment (S4); subdivided into subsegment (4a) and (4b) Right pa of the liver 1. Right medial division 1. Right anteromedial segment (S5) 2. Right posteromedial segment (S8) 2. Right lateral division 1. Right anterolateral segment (S6) 2. Right posterolateral segment (S7)

So the answer to this question is B


Instruction:
Hypercalcemia is caused by all EXCEPT :
A. Loop diuretics
B. Lithium
C. Vitamin D intoxication
D. Thiazides

Response:
LOOP loses CALCIUM. Loop diuretics cause hypocalcemia by more excretion whereas thiazides cause hyperkalemia by decreasing its excretion.

So the answer to this question is A


Instruction:
Modified Ziehl-neelsen staining is used for:
A. Mycobacterium tuberculosis
B. Mycobacterium bovis
C. Nocardia
D. All of the above

Response:
Ans. is. 'd' i. e., All of the above

So the answer to this question is D


Instruction:
Calcitonin is used in all of the following conditions, EXCEPT:
A. Hypervitaminosis D
B. Postmenopausal osteoporosis
C. Paget's disease
D. Fanconi syndrome

Response:
The ability of calcitonin to block bone resorption and lower serum calcium makes it a useful drug for the treatment of Paget's disease, hypercalcemia, and osteoporosis. Fanconi syndrome is a disease of the proximal renal tubules of the kidney in which glucose, amino acids, uric acid, phosphate and bicarbonate are passed into the urine, instead of being reabsorbed. Treatment of children with Fanconi syndrome mainly consists of replacement of substances lost in the urine (mainly fluid and bicarbonate). Calcitonin is not indicated in this disease. In paget's disease bisphosphonates and calcitonin decrease the elevated biochemical markers of bone turnover, such as plasma alkaline phosphatase activity and urinary excretion of hydroxyproline. Ref: Bikle D.D. (2012). Chapter 42. Agents that Affect Bone Mineral Homeostasis. In B.G. Katzung, S.B. Masters, A.J. Trevor (Eds), Basic & Clinical Pharmacology, 12e.

So the answer to this question is D


Instruction:
A patient presents with history of episodic painful edema of face and larynx. Which of the following is likely to be deficient –
A. Complement C3
B. Complement C5
C. C1 Esterase Inhibitor
D. Properidin

Response:
Recurrent episodes of edema of face and larynx suggest the diagnosis of hereditory angioneurotic edema caused by deficiency of C1 esterase inhibitor.

So the answer to this question is C


Instruction:
Following a sudden impact in an accident, the 34-year-old race car driver becomes unconscious and is admitted to the hospital. A CT scan is performed, and a right space-occupying lesion is noted (Fig. 11-1). What is the most likely diagnosis?
A. Corpus callosum injury
B. Pituitary apoplexia
C. Acute subdural hematoma
D. Acute epidural hematoma

Response:
Acute subdural hematomas occur most commonly when violent accelerations or deceleration injuries of the head cause tearing of the bridging veins within the subdural potential space. They generally imply a much more severe injury to the brain itself than in the case of their epidural counterpa. For this reason, they are associated with cerebral contusions in over 30% of cases.

So the answer to this question is C


Instruction:
Great vein of Galen is formed by
A. Two internal cerebral veins
B. Superficial middle cerebral vein & deep middle cerebral vein
C. Anterior cerebral vein & basal vein
D. Internal cerebral veins & straight sinus

Response:
(A) Two internal cerebral veins # GALEN'S VEINS: The internal cerebral veins; the great vein of Galen is formed by their union.> Anterior cerebral vein also known as Galen's vein.> Great cerebral vein or Great vein of Galen is formed by union of two internal cerebral veins. It terminates in the straight sinus.

So the answer to this question is A


Instruction:
All of the following nerves are posteriorly related to the kidney, EXCEPT?
A. Subcostal Nerve
B. Iliohypogastric Nerve
C. Lateral Cutaneous Nerve of the thigh
D. Ilioinguinal Nerve

Response:
Posterior to the kidney, between the quadratus lumborum muscle and the posterior surface of the kidney, are the subcostal nerve (T12) and vessels, the iliohypogastric and the ilioinguinal nerves (L1) passing inferolaterally. The lateral cutaneous nerve of the thigh is not posteriorly related to the kidney.

So the answer to this question is C


Instruction:
Which of the following is a serious and characteristic adverse effect of Metformin
A. Lactic acidosis
B. Weight gain
C. Hypoglycemia
D. Dilutional hyponatremia

Response:
Ans: A. Lactic acidosis [Ref: KDT 7h/e p. 453).Two impoant adverse effects of these agents are lactic acidosis (more with phenformin) and vitamin B12 deficiency (more common with metformin).More common side effects are abdominal pain, anorexia, metallic taste, mild diarrhea and tiredness.

So the answer to this question is A


Instruction:
After overnight fasing, levels of glucosetranpoers reduced in
A. Brain cells
B. RBCs
C. Adipocytes
D. Hepatocytes

Response:
Adipocytes (Ref: Harper 28/e, p138, 171 (26/e p160, 232); Lippincott Biochem 3/e p310; Ganong23/e p320] Overnight fasting causes insulin levels to fall. This decreases the number of glucose transpoers which are insulin dependent. Insulin dependent glucose transpoers are GLUT-4, present in adipocytes, skeletal and cardiac muscles. Rest of the glucose transpoers are not insulin dependent. Harper writes-"Glucose uptake into muscle and adipose tissue is controlled by insulin, which is secreted by the B islet cells of the pancreas in response to an increased concentration of glucose in the poal blood. An early response to insulin in muscle and adipose tissue is the migration of glucose transpoer vesicles to the cell surface, exposing active glucose transpoers (GLUT 4). These insulin-sensitive tissues will only take up glucose from the blood stream to any significant extent in the presence of the hormone. As insulin secretion falls in the starved state, so the transpoers are internalized again, reducing glucose uptake" Glucose Transpoers Glucose enters cells either by facilitated diffusion or by secondary active transpo with No+ (in the intestine and kidneys) Glucose transpoers are a family of membrane proteins that are responsible for facilitated diffusion of glucose across cell membranes. They differ from and have no homology with the sodium-dependent glucose transpoers, SGLT 1 and SGLT 2, responsible for the secondary active transpo of glucose in the intestine and renal tubules. Seven different glucose transpoers, named GLUT 1-7 in order of discovery. have been characterized. Each transpoer appears to have evolved for special tasks. GLUT 4 is the transpoer in muscle and adipose tissue that is stimulated by insulin. Glucose Transpoers in Mammals Function Major Sites of Expression Facilitated diffusion GLUT 1 Basal glucose uptake Brain, kidney, colon, placenta, erythrocytes GLUT 2 B-cell glucose sensor; transpo out of intestinal and renal epithelial cells B cells of pancreatic islets, liver, small intestine, kidneys GLUT 3 Basal glucose uptake Brain, placenta, kidneys GLUT 4 Insulin-stimulated glucose uptake Skeletal and cardiac muscle, adipose tissue GLUT 5 Fructose transpo Jejunum, sperm GLUT 6 None Pseudogene GLUT 7 Glucose 6-phosphate transpoer in endoplasmic reticulurn Liver, ? other tissues Secondary active transpo (Nal? glucose cotranspo) SGLT 1 Absorption of glucose Small intestine, renal tubules SGLT 2 Absorption of glucose Renal tubules

So the answer to this question is C


Instruction:
A 1-year-old child, on a routine well child visit, was discovered to have cataract formation in both eyes. Blood test demonstrated elevated galactose and galactitol levels. In order to determine which enzyme might be defective in the child, which intracellular metabolite should be measured?
A. Galactose
B. Fructose
C. Glucose
D. Galactose-1-phosphate

Response:
The child has a form of galactosemia. The elevated galactose enters the lens of the eye, is reduced to galactitol, and is trapped. The difference in osmotic pressure across the lens of the eye leads to cataract formation. Galactose is phosphorylated by galactokinase to galactose-1-phosphate, which reacts with UDP-glucose in a reaction catalyzed by galactose-1-phosphate uridylyl transferase to form UDP-galactose and glucose-1-phosphate. An epimerase converts UDP-galactose to UDP-glucose. Deficiencies in either galactokinase (nonclassical) or galactose-1-phosphate uridylyl transferase (classical) result in galactosemia, with elevated levels of galactose and galactitol (reduced galactose) in the blood. An intracellular measurement of galactose-1-phosphate can allow a definitive diagnosis to be obtained (such levels would be nonexistent if the defect were in galactokinase, and the levels would be greatly elevated if the galactose-1-phosphate uridylyl transferase enzyme were defective).

So the answer to this question is D


Instruction:
In &;Hotair oven&; for holding period of one hour, temperature required is -
A. 100o C
B. 120o C
C. 140o C
D. 160 o C

Response:
Temperature and time: 160 o C for 2 hours (holding time ) is required for sterilization However alternative temperatures and holding time include 170degC for 1 hour and 180degC for 30 minutes Ref: a textbook of Microbiology Dr.C.P.Baveja 4th Edition pg no:30

So the answer to this question is D


Instruction:
A 33-year-old male is admitted to the hospital after a violent, multiple car collision. His blood pressure is 89/39 mm Hg, and a central venous line is ordered to be placed. Which of the following structures is used as a landmark to place the tip of the catheter of the central venous line?
A. Carina
B. Subclan aery
C. Superior vena cava
D. Right atrium

Response:
The carina can easily be seen in radiograph. The carina is at the level of T4-5 (plane associated with the sternal angle of Louis). This landmark is commonly used to guide the placement of a central venous line.

So the answer to this question is A


Instruction:
Which among the following is not an androgen receptor blocker?
A. Finasteride
B. Cyproterone
C. Flutamide
D. None

Response:
Finasteride is a 5alpha-reductase inhibitor used for BPH, Prostate cancer Flutamide and Cyproterone acetate are AR antagonists. Finasteride is used to shrink an enlarged prostate (benign prostatic hyperplasia or BPH) in adult men. It may be used alone or taken in combination with other medications to reduce symptoms of BPH and may also reduce the need for surgery. (Ref.Essentials of medical pharmacology TD Tripathi 7th edition page no.302)

So the answer to this question is A


Instruction:
True about Pneumocystis carinii (Pneumocystis jerovecii) is the following EXCEPT
A. Respiratory alkalosis
B. | diffusion capacity
C. Cyanosis
D. | arterial O2 gradient (PAO2 - PaO2)

Response:
Ans. d (| arterial O2 gradient (PAO2-PaO2)) (Ref: Harrison's Internal medicine 17th edition, p. 1267)ABG in PCP:# | alveolar arterial O2 gradient (Normal < 15 mm H2O 20-30 )# | PaO2 When,When,PAO2 - PaO2 < 35 mm Hg:- PAO2 - PaO2 > 35 mm Hg and PaO2 < 60 mm Hg:- Mild infection- Severe infection- No need to add steroids- Add steroid for 14-21 days.Pneumocystis JiroveciiFungus - Ribosomal genetic make-up and RNA resembles to that of fungus, but lacks mycolic acid in cell-wall.# When it occurs? -CD4 count < 200/mcc in HIV patients# When to suspect? -Fever > 2 weeksDry cough > 2 weeksPresence of oropharyngeal/oesophageal Candida# Clinical features: Dry cough, Dyspnea, Tachycardia, Fever, Cyanosis# RS examination: Auscultation reveals only few abnormalities# CXR: Most common reticulonodular infiltrates starting at hillum involving MZ and LZ with sparing of apex and CP angle (perihilar ground-glass haziness is earliest radiological change in PCP).# Rarest - pleural effusion.# Pneumothroax especially receiving (5-10% cases) aerosolized pentamidine.# Diffusion capacity of lung# Other sites of P Jirovecii infection- Lymph node (most common extrapulmonary site)0- Spleen- Liver- Bone Marrow# Treatment- DOC - Septran- Other Drugs: Dapsone, Pentamidine, Clindamycin and Atovaquone.Summary - Pneumocystis jirovecii:# Yeast (originally classified as protozoan). Inhaled. Most inf asymptomatic. Immunosuppression (AIDS) pre- disposes.# Causes Pneumocystis pneumonia (PCP), a diffuse interstitial pneumonia.# Diffuse, bilateral CXR appearance. Diagnosed by lung biopsy or lavage. Disc-shaped yeast forms on meth- enamine silver stain of lung tissue.# Treatment/prophylaxis: TMP-SMX, pentamidine, dapsone (prophylaxis only), atovaquone (prophylaxis only). Start prophylaxis when CD4 count drops < 200 cells/mm3 in HIV patients.Also Know:Conditions where diffusion capacity of Lung increasedEarly LVFIntra-alveolar hemorrhage

So the answer to this question is D


Instruction:
An infant presented with erythematous lesions on cheek, extensor aspect of limbs, mother has history of bronchial asthma, the probable diagnosis is
A. Air borne contact dermatitis
B. Atopic dermatitis
C. Seborraehic dermatitis
D. Infectious eczematoid dermatitis

Response:
B i.e. Atopic Dermatitis

So the answer to this question is B


Instruction:
Half-life of basiliximab is:-
A. 7 days
B. 7 hours
C. 15 days
D. 24 hours

Response:
Half life of basiliximab is 7 days.Basiliximab* This is anti CD-25 antibody with higher affinity for the IL-2 receptor, but shoer plasma t 1/2 (1 week).* It is used to prevent renal and other transplant rejection reaction.* It can cause anaphylactic reactions and promote oppounistic infection.

So the answer to this question is A


Instruction:
Lactose content of breast milk per 100 ml is
A. 3.5 gm
B. 4.5 gm
C. 6 gm
D. 7 gm

Response:
Lactose content of breast milk per 100 ml is 7.0 g. Ref: Table 5.4, Page 97; Ghai essential pediatrics; 6th edition

So the answer to this question is D


Instruction:
Drugs indicated in MRSA infection are all except: March 2012
A. Vancomycin
B. Teicoplanin
C. Linezolid
D. Imipenem

Response:
Ans: D i.e. Imipenem Drugs and indications Vancomycin is the preferred surgical prophylactic in MRSA prevalent areas and in penicillin allergic patients Teicoplanin is indicated in enterococcal endocarditis (along with gentamicin); MRSA and penicillin resistant streptococcal infections, osteomyelitis and as alternative to vancomycin Linezolid is active against MRSA and some VRSA, VRE, penicillin-resistant Strep. Pyogenes, Strep. viridians and Strep. Pneumonia, M. tuberculosis, Corynebacterium, Listeria, Clostridia and Bact. Fragilis Imipenem (with cilastatin) has proved effective in a wide range of serious hospital acquired infections including those in neutropenic, cancer and AIDS patients

So the answer to this question is D


Instruction:
Which of the following is not a prefabricated pontic:
A. Long pin facing.
B. Pontip.
C. Perel pontic.
D. Sanitary facing.

Response:
Example of prefabricated pontics are:
Trupontic, Long pin facing,
Flatback, Sanitary facings, Reverse pin facing, Pontips.

So the answer to this question is C


Instruction:
A 78-year-old man has had increasing dyspnea without cough or increased sputum production for the past 4 months. On physical examination, he is afebrile. Breath sounds are reduced in all lung fields. A chest CT scan shows a dense, brightly attenuated pleural mass encasing most of the left lung. Microscopic examination of a pleural biopsy specimen shows spindle and cuboidal cells that invade adipose tissue. Inhalation of which of the following pollutants is the most likely factor in the pathogenesis of this mass?
A. Asbestos
B. Bird dust
C. Coal dust
D. Cotton fibers

Response:
Malignant mesothelioma is a rare tumor even in individuals with a history of asbestos exposure. The tumor may appear decades after exposure, and is not related to amount or length of exposure. Bronchogenic carcinoma is more common in individuals with asbestos exposure, particularly when there is a history of smoking. Bird dust inhalation can lead to hypersensitivity pneumonitis. Coal dust inhalation can lead to marked anthracosis, but without a significant risk of lung cancer. Inhalation of cotton fibers (byssinosis) leads to symptoms resembling asthma related to bronchoconstriction. Ozone and nitrogen oxides in smog can cause acute respiratory discomfort, but are not known to be promoters of neoplasia. Silicosis is typified by interstitial fibrosis and causes a slight increase in the risk of bronchogenic carcinoma.

So the answer to this question is A


Instruction:
Gentian violet colouration of gram positive bacteria is due to ?
A. Peptidoglycan
B. Capsule
C. Cell membrane
D. None of the above

Response:
Ans. is 'a' i.e., Peptidoglycan GRAM STAINING The gram stain differentiates bacteria into two broad groups :? Gram positive - Resist decolourization and retain primary stain, appear violet. Gram negative - Decolourised by organic solvents, and therefore, take the counterstain, appearing red.

So the answer to this question is A


Instruction:
True regarding ondansetron are all of the following except: March 2010
A. Antiemetic of choice for chemotherapy induced vomiting
B. 5-HT3 receptor antagonist
C. Significant drug interactions
D. May have 5-HT4 antagonism

Response:
Ans. C: Significant Drug Interactions Ondansetron is the prototype of new class of antiemetic drugs developed to control chemotherapy/radiotherapy induced vomiting, and later found to be highly effective in postoperative nausea and vomiting as well. It blocks the action of 5-HT through 5-HT3 receptors on vagal afferents in the GIT as well as in NTS and CTZ. A minor 5-HT4 antagonistic action has also been shown. It is hydroxylated by CYP 1A2, 2D6 and 3A, but no clinically significant drug interactions have been noted.

So the answer to this question is C


Instruction:
Which is a product of purine metabolism?
A. Uric acid
B. Urea
C. Xanthine
D. Beta alanine

Response:
Uric acid is the common end product of purine catabolism. However, in non-primate mammals, Allantoin is the end product of purine catabolism as an extra enzyme, Uricase is present. OTHER OPTIONS: OPTION B: The end products of pyrimidine catabolism are NH3, CO2 & beta alanine (for Cytosine, Uracil) Option C - Xanthine is an intermediate of catabolism of purines. Xanthine gets conveed to uric acid by Xanthine Oxidase.

So the answer to this question is A


Instruction:
Metal fume fever is seen with
A. Mercury
B. Lead
C. Zinc
D. Thallium

Response:
Metal fume fever This is due to inhalation of zinc oxide fumes. It is also called smelter's shakes, zinc shakes, brass chills or Monday morning fever. Ref: FORENSIC MEDICINE AND TOXICOLOGY Dr PC IGNATIUS THIRD EDITION PAGE 430

So the answer to this question is C


Instruction:
True about LGV ?
A. Bleeding ulcer in 1" stage
B. Bubos in 1" stage
C. Groove's sign in rd stage
D. Proctocolitis in 2nd stage

Response:
Ans. is 'c' i.e., Groove's sign in 2' stageStages of LGV are :?First stage (Primary LGV) :- Self limited, Single, asymptomatic, painless, nonbleeding genital ulcer.Secondary stage : - Painful inguinal lymphadenopathy (Remember Ulcer is pain less but lymphodenopathy is tender & painful).Swollen lymph nodes coalesce to form bubos, i.e., matted lymph nodes. Buboes may rupture to form discharging sinus. Groove's sign - Enlarge lymph nodes both above and below inguinal ligament.Teiary LGV (genitorectal syndrome) : - Characterized by proctocolitis.

So the answer to this question is C


Instruction:
Tubular vision is seen in
A. Retinitis pigmentosa
B. Bread crumb cataract
C. Balint syndrome
D. central serous retinopathy

Response:
Ans. (a) Retinitis pigmentosa

So the answer to this question is A


Instruction:
The primary cause for the development of shock following hemorrhage is
A. Marked vasodilation
B. Decreased blood volume
C. Inadequate output by the hea
D. Obstruction to blood flow

Response:
Hemorrhage is the most impoant cause of hypovolemic shock. There will be reduction in the venous return leading to a fall in cardiac output and BP. The severity of shock depends upon the rapidity of blood loss. Ref: Textbook of medical physiology by N Geetha, 3rd edition, page no. 204

So the answer to this question is B


Instruction:
Memory impairment is most likely to occur in:
A. Downs Syndrome
B. Alkaptonuria
C. Attention deficit disorder
D. Conduct disorder

Response:
Answer is A (Down's syndrome) Down Syndrome is associated with memory impairment. Adults with trisomy 21 (Down's) consistently develop the neuropathologic hallmark's of Alzheimer's disease of they survive beyond 40 years of age - Harrison 's The most noble aspect is down's syndrome is marked mental retardation, relatively poor expressive language and impaired verbal sho term memory' - Child and Adolescent Psychiatry

So the answer to this question is A


Instruction:
ADH formation site is:
A. Pituitary
B. Hypothalamus
C. Pineal
D. None

Response:
Supraoptic and paraventricular nuclei synthesize ADH hormone, which travel along the axons of these neurons and is secreted from posterior pituitary.

So the answer to this question is B


Instruction:
Which of following Wibrios' is most commonly associated with ear infections -
A. V. Alginolyticus
B. V. Parahemolyticus
C. V. Vulnifcus
D. Vibrio flulis

Response:
Ans. is 'a' i.e., V. Alginolyticus Vibrio Alginolyticus is an oppounistic pathogen (Halaphilic vibrio) that is primarily associated with wound and ear infections and occasionally eye infections. Features of Selected Noncholera Vibrios Organism Vehicle or Activity Host at Risk Syndrome V parahaemolyticus Shellfish, Seawater Seawater Normal Normal Gastroenteritis Wound infection Non- 01 V cholerae Shellfish, travel Seawater Normal Normal Gastroenteritis Wound infection otitis media V vulnificus Shellfish Immunosuppressed cellulitis Sepsis, secondary Seawater Normal Wound infection cellulitis V. alginolyticus Seawater Seawater Normal Burned, other immunosuppressed Wound infection cellulitis, otitis Sepsis See in above table, otitis media is caused by Non-01 V. cholerae and V. alginolyticus.

So the answer to this question is A


Instruction:
Posterior cerebral aery supplies all except
A. Pons
B. Thalamus
C. Midbrain
D. Lentiform nucleus

Response:
PCA SuppliesOccipital lobeTemporal lobeThalamusMidbrainLentiform nucleusPineal bodyMedial geniculate bodyLateral and 3rd ventricle

So the answer to this question is A


Instruction:
Which of the following is not a feature of mediastinal syndrome
A. Dysphagia
B. Intercostal neuralgia
C. Hoarseness of voice
D. Engorgement of Veins in lower half of body

Response:
Mediastinal syndrome other name for SVC syndrome or It involves veins of upper half of body.

So the answer to this question is D


Instruction:
A terminally ill patient with Ca pancreas is on IV Diamorphine 10 mg/day in a cancer institute in Kerala. He is being discharged from the hospital and would like to conve it to Oral Morphine. What is the approximate dose of oral Morphine needed in this patient?
A. 20 mg
B. 30 mg
C. 40 mg
D. 5 mg

Response:
The potency of some drugs in relation to morphine is given below IV morphine - potency 2 x oral morphine IV diamorphine - potency 3 x oral morphine Oxycodone P.O - potency 2 x oral morphine Oxycodone IV - potency 3 x oral morphine Ref: Oxford handbook of clinical specialities, 8th Edition , Page 535

So the answer to this question is B


Instruction:
Sine wave pattern on ECG is done when serum potassium exceeds mEq/dl?
A. > 6 mEq/dl
B. > 7 mEq/dl
C. > 8 mEq/dl
D. > 10 mEq/dl

Response:
Answer- C. > 8 mEq/dlSine wave pattern- > 8 mEq/dl

So the answer to this question is C


Instruction:
Which of the following statements is true regarding menorrhagia?
A. Heavy and regularly timed episodes of bleeding
B. Heavy and irregularly timed episodes of bleeding
C. Amount of bleeding greater than 50 ml
D. Duration of bleeding greater than 3 days

Response:
Menorrhagia is defined as heavy but regular menstrual bleeding. Menorrhagia denotes regularly timed episodes of bleeding that are excessive in amount (>80mL) and/or duration of flow (>7 days) According to the recent update, the term heavy menstrual bleeding has replaced menorrhagia. Hemoglobin concentration below 12 g/dL increase the chance of identifying women with heavy menstrual bleeding. A normal level, however, doesnot exclude heavy menstrual bleeding, as many women with clinically significant bleeding have normal values. All patients should undergo a pelvic examination and, if the menstrual pattern has changed substantially or if anemia is present, a vaginal sonography should be carried out as the most impoant supplemental examination. Vaginal sonography combined with an endometrial biopsy is a reliable method for diagnosing endometrial hyperplasia or carcinoma, but it is insufficient for diagnosing endometrial polyps and fibroids; these can be diagnosed more reliably by sonohysterography or hysteroscopy. Appropriate treatment considerably improves the quality of life of the patients. Non-steroidal anti-inflammatory drugs and tranexamic acid reduce menstrual blood loss by 20-60% Drug treatment should be used and evaluated before surgical interventions are considered. Ref: Shaw&;s textbook of Gynaecology 17th edition Pgno: 122

So the answer to this question is A


Instruction:
Which of following are the risk factors for inguinal hernia?
A. COPD
B. BPH
C. Cigarette smoking
D. All of the above

Response:
Risk factors for hernia Weakness of abdominal muscles Increased intra-abdominal pressure Patent processus vaginalis Patent canal of nuck causing indirect inguinal hernia in females Connective tissue disorders like Ehlers Danlos Syndrome. Congenital conditions like Exstrophy of bladder, Prune Belly syndrome Advancing age Chronic debilitating disease Defective collagen synthesis Previous right lower quadrant incision Cigarette smoking Chronic cough (Bronchitis, tuberculosis) Chronic obstructive pulmonary disease Obesity Chronic constipation with straining at stool Enlarged prostate with straining at micturition Pregnancy Cirrhosis with ascites Heavy weight lifting Chronic ambulatory peritoneal dialysis Intra-abdominal tumors Chronically enlarged pelvic organs

So the answer to this question is D


Instruction:
Drugs used for recurrent depressive episodes -a) Imipramineb) Carbamazepine c) Fluoxetined) Na valproate
A. a
B. bc
C. ac
D. b

Response:
Amongst the given options, Imipramine and Fluoxetine are antidepressants.
Carbamazepine and Valproate are used for prophylaxis of BPD (not for recurrent unipolar depression).

So the answer to this question is C


Instruction:
True about HIV
A. Not transmitted through semen
B. More chances of transmission during LSCS than normal labour
C. More infectious than hepatitis B
D. Male to female transmission > female to male

Response:
(D) Male to female transmission > female to male > One reason for high incidence of HIV infection in male homosexuals is because of large numbers of sexual partners they have> Homosexual and bisexual men are more affected more than heterosexuals.> Male to female propagation is more because of large receptive area in females.> More chances of transmission during normal labuor, therefore LSCS is perferred.> Hepatitis B is more infectious than HIV.

So the answer to this question is D


Instruction:
Terminal axillary draining lymph nodes -
A. Anterior
B. Posterior
C. Scapular
D. Apical

Response:
Ans. is 'd' i.e., Apical o The apical group of axillary lymph node receives lymph from anterior (pectoral), posterior (subscapular), central and lateral (humeral) group of axillary lymph nodes. Hence apical group of lymph node is the terminal group of axillary lymph node.Axillary lymph nodeRelated vesselsAnterior (pectoral)Along lateral thoracic vesselsPosterior (scapular)Along subscapular vesselsLateralMedial to axillary veinApical (terminal or infraclavicular)Along axillary vessels

So the answer to this question is D


Instruction:
Diagnosis of Cervical Incompetence is established by use of which size Hegar&;s dilator?
A. 4
B. 6
C. 8
D. 10

Response:
Ans. is 'C' i.e. 8 o Diagnosis in pregnant state: 1) History: Painless second trimester abortion in previous gestation. 2) Shirodkar's digital palpation test: # In suspected cases (those with a previous abortion), do a weekly per vaginal examination. A finger introduced upto the internal os and if found open pass it a little beyond and withdraw the finger. While withdrawing feel for the sphincter - like tone of the internal os; if the os tends to close down with sphincter like or snapping action, the os is considered competent and no cerclage is necessary. This is a highly subjective test. 3) Cervical sonography: # Funelling of the cervix with changes in the form of Y, V, U. # Cervical length < 2.5 cm. # Funnelling of the internal os > 1 cm # Speculum examination: Detection of dilatation of internal os with herniation of the membranes. # Cervical index = (Funnel length + 1) / (endocervical length). # The "+1" allows an index to be calculated when funelling is absent. o Predictors of preterm birth: # Cervical index >0.52. # Cervical length < 18 mm # Funnel length > 9 mm. # Funnel width > 6 mm. o Diagnosis in non - pregnant state: Diagnosis of cervical incompetence in the non-pregnant female is done by Passage of no. 8 hegars dilator during the premenstrual phase of the menstrual cycle. 1) Passage without resistance and pain of No. 8 Hegarrs dilator. 2) Cervicogram : shows funnel shaped cervix. 3) Intracervical balloon studies: Pass a balloon (e.g. Foley's) into cervical canal and inflate it with a radio - opaque dye and then take x - rays after applying due traction. Pear shaped shadow of balloon due to funneling of cervical canal indicates incompetent os. 4) Traction studies : Pass foley's catheter through internal os of cervix and inflate it with 5 - 10 ml and apply traction. If traction of 600 gm or less is enough to pull out the bulb through the os, then the diagnosis is certain. 5) Bromen papaine test : This dye selectively dilates the cervico - uterine junction. After injecting the dye (locally into the cervix and intramuscularly.

So the answer to this question is C


Instruction:
The drug for OPD analgesia is -
A. Morphine
B. Pethidine
C. Fentanyl
D. Alfentanil

Response:
Ans. is 'd' i.e., Alfentanil Drugs useful for day care surgery o Propofol o Sevoflurane, desflurane & isoflurane o Midazolam o Mivacurium o Alfantanil

So the answer to this question is D


Instruction:
A/E one results in increased secretion of Renin
A. Renal ischaemia
B. Decreased amount of Na+ in DCT
C. Decreased amount of Na+ in PCT
D. Narrowing of afferent aerioles

Response:
C i.e. Decreased amount of Na+ in PCT

So the answer to this question is C


Instruction:
What is the most likely diagnosis of this penile lesion found in a 54-year-old man with lymphadenopathy and weight loss?
A. Squamous-cell carcinoma
B. Syphilis
C. Extramammary Paget disease
D. Condyloma acuminatum

Response:
Answer A. Squamous-cell carcinomaSquamous-cell carcinoma is the correct answer. Human papillomavirus (HPV) is a common risk factor for squamous-cell carcinoma of the penis, as is the case here. HPV can also cause cervical cancer in women. The presence of metastasis to the inguinal lymph nodes is a poor prognostic factor for survival. Primary prevention, education, and vaccination are important to reduce the occurrence of HPV infection and its associated cancers.

So the answer to this question is A


Instruction:
Reversal of SMA and SMV relationship is seen in?
A. Midgut Volvulus
B. Duodenal Atresia
C. Annular pancreas
D. Ileal Atresia

Response:
Figure: Inversion in the SMA/SMV relationship with the SMA on the right and the SMV on the left. Figure: Whirlpool sign with twisting of mesentery and mesenteric vessels, is seen in Malrotation/ MidGut Volvulus.

So the answer to this question is A


Instruction:
Corneal curvature measured by -
A. Keratometry
B. Pachymetry
C. Gonioscopy
D. Perimetry

Response:
Ans. is 'a' i.e., Keratometry o The 'Keratometry' or 'ophthalmometry' is an objective method of estimating the corneal astigmatism by measuring the curvature of central cornea.o Gonioscopy - Biomicroscopic examination of the angle of anterior chamber,o Perimetry - Procedure of estimating the extent of visual fields.o Tonography - Non - invasive technique for determining the facility of aqeous outflow (C - value).

So the answer to this question is A


Instruction:
Einthovens law -
A. I+ III= II
B. I -III=II
C. I+II+III= 0
D. I+III=avL

Response:
Ans is 'a' i.e., I + III = II o Einthoven's triangle is an equilateral triangle formed by two shoulder and the pubis with the heart is in the center as moving dipole.o For convenience Eithoven's triangle consider right arm (RA), left arm (LA), and Left leg (LL)Q to which electrodes are connected.o An electrode connected to all the three comers of Eithoven's triangle will always be at zero potentials regardless of the direction of the cardiac vector.o Such an electrode is called the indifferent electrode (or the central terminal of wiison)Q.o The indifferent electrode is connected to RA, LA and LL through high (5000 Ohm) resistance.ECG Leadso An ECG is recorded by picking up potentials from two parts of the body. The two sites selected for this purpose constitue the ECG leads. Usually ECG is recoreded by 12 leads (12 lead ECG): - Three bipolar (standard) limb leads, three Unipolar limb leads and six unipolar chest leads.Bipolar (standard) limb leads or Einthoven's leado Bipolar recording involves recording of potential difference between two different sites on the body using two electrodes, one serving as the reference and the other called the exploring or active electrode. These leads are:-Lead I: - Between right arm (negative/reference electrode) and left arm (positive or active or exploring electrode).LeadII: - Between right arm (negative or reference electrode) and left leg (positive electrode).Lead III: - Between left arm (negative electrode) and left leg (positive electrode).o The direction in wrhich positive electrode points is called the axis of lead. If the direction of the lead axis is similar to the direction of the flow of current in the heart, the recorded deflection is positive (upward). Hence electrodes are arranged in such a way that most of the deflections in the ECG are positive,o Einthoven's law Q: - This law states that if the electrical potential of any of the three bipolar limb leads are known at any given instant, the third one can be determined mathematically by simply summing the first two (but note that the positive and negative signs of different leads must be observed when making this summation). Thus the sum of the voltages in leads I and III equals the voltage in lead IIQ.Unipolar limb leadso Unipolar recording involves recording the potential difference between an active electrode and an indifference electrode. Thus, unipolar limb leads and chest leads employ an active electrode and an indifference electrode. The indifferent electrode is connected to left arm, right arm and left leg through a very high resistance (has been explained). Because of the high resistance, the indifferent electrode stays at an almost constant (zero) potential. Since the leads measure the potential difference between two electrodes, one of which is at constant voltage (indifferent), it essentially measures the potential at the other (active) electrode. There are three unipolar limb leads:-o VR : Between right arm and an indifferent electrode,o VL : Between left arm and an indifferent electrode,o VF: Between left leg and an indifferent electrode.o These unipolar limb leads are not much used in practice because augmented limb leads which provide a magnified version of the same information; therefore, unipolar augmented limb leads are used. Augmented limb leads are recordings between one limb and two other limbs12, i.e., the positive (active) electrode is connected to one limb and the negative electrode to the other two through high resistances. This results in an increase in the voltages recorded in the ECG That is why these leads are called augmented limb leads. There are three augmented limb leads: -aVR : - Between right arm (positive electrode) and left arm + left leg (negative or refrence electrode).aVL : - Between left arm (positive electrode) and right arm + Left leg (negative electrode).aVF : - Between left leg (positive electrode) and right arm + left arm (negative electrode).Unipolar chest leadsQo These leads emply an exploring (active or positive) electrode on the chest and an indifferent (reference or negative) electrode. The indifferent electrode is connected to the right arm, left arm and left leg through high resistances. The position of chest electrode (positive electrode) is as follows (VI - V6Q): -V1: In the right 4th intercostal space at the right border of the sternum.V2: In the left 4th intercostal space at the left border of sternum.V3: At midpoint between V2 and V4.V4: In the left 5th intercostal space in the midclavicuiar line.V5: In the left 5th intercostal space at the anterior axillary line.V6: In the left 5th intercostal space at the mid-axillary line.

So the answer to this question is A


Instruction:
Urea breath test is done in:
A. Ketoacidosis
B. H. pylori infection
C. Kidney disease
D. Liver disease

Response:
The urea breath test is a rapid diagnostic procedure used to identify infections by Helicobacter pylori, a spiral bacterium implicated in gastritis, gastric ulcer, and peptic ulcer disease.
It is based upon the ability of H. pylori to convert urea to ammonia and carbon dioxide.

So the answer to this question is B


Instruction:
All are used in heroin poisoning EXCEPT:
A. Cloni di ne
B. Buprenorphine
C. Pentazocine
D. Haloperidol

Response:
D i.e. Haloperidol

So the answer to this question is D


Instruction:
Inflammasome is formed in
A. Necrosis
B. Apoptosis
C. Necroptosis
D. Pyroptosis

Response:
Ans. (d) Pyroptosis(Ref: Robbins 9th/pg 58-59)Pyroptosis: Promotes the activation of inflammasomewhich activates Caspase-1Q which releases biologically active form of IL^-1 from the precursor.

So the answer to this question is D


Instruction:
Which of the tumors are unique to pregnancy
A. Luteoma
B. Serous cystadenoma
C. Mucinous cystadenoma
D. Teratoma

Response:
Pregnancy luteomas are rare, nonneoplastic lesions of the ovary thought to be caused by the hormonal effects of pregnancy. Most of these patients are asymptomatic with the ovarian enlargement being incidentally discovered during imaging or surgery. Some patients develop hirsutism or virilization during late pregnancy. Luteomas spontaneously regress postpaum. It may be a diagnostic and management challenge as it can mimic the presentation of malignant ovarian tumors.

So the answer to this question is A


Instruction:
Von-Rosen's splint used in:
A. Congenital dislocation of hip joint
B. Congenital taloequinovarus
C. Tibia vera
D. Congenital coxa-vara

Response:
(Congenital dislocation of hip joint) (203-M)Congenital dislocation of hip(a) Plaster cast - A frog leg (Lorenz cast) or Bachelor cast(b) Splint - Von Rosen's splint

So the answer to this question is A


Instruction:
'Time out' is useful in the treatment of:
A. Temper Tantrums
B. Pica
C. Anorexia nervosa
D. Nocturnal enuresis

Response:
Time-out- behavioral modification that involves temporarily separating child from an environment where an unacceptable behavior has occurred. The goal is to remove that child from an enriched, enjoyable environment- lead to extinction of the offending behavior. Distraction and 'time-out' strategies are useful in treatment of Temper Tantrums.

So the answer to this question is A


Instruction:
Which organism is associated with causation of urinary bladder cancer -
A. E. coli
B. Paragonimus
C. Schistosoma
D. Clonorchis

Response:
Ans. is 'c' i.e., SchistosomaSchistosomes* These flukes (trematodes) reside in blood of venous plexus at different sites.* Three important species are:-i) S. haematobium : Veins of vesical & pelvic plexuses, less commonly in portal vein and its mesentric branch.ii) S. mansani: Mesentric veins draining sigmoido-rectal region (inferior mesentric veins).iii) S. Japonicum : Mesentric veins draining ilio-caecal region (superior mesentric vein).* Schistosomes are transmitted by Snails (Secondary host) to man (definitive host). Infective form in snail is cercaria.* Eggs of S. hematobium have terminal spine. Eggs of S.mansoni have lateral spine and eggs of S.japonicum have lateral knob.* S. hematobium causes bilharziasis (urinary schistosomiasis) which is characterized by painless terminal hematuria.* S. japonicum and S.mansoni cause Katayama fever. Rarerly, S. hematobium can also cause Katayama fever.* Fairly's test is an intradermal skin test with cercarial antigen. It is an allergic reaction, positive in all the varieties of schistosomiasis.* Schistosoma hematobium is a well known cause of urinary bladder cancer.

So the answer to this question is C


Instruction:
A 65-year-old man presents to your clinic after falling several times in the past few months. On examination, his most notable findings are an unstable, wide-based gait and marked retropulsion. He does have bradykinesia with masked facies and dysahria. You find no evidence of cogwheeling or resting tremor. His symptoms do not respond to levodopa/carbidopa. During the next few months, his eye movements are notable for the slowing of veical saccades and fast phases. What is the most likely diagnosis?
A. L-DOPA resistant PD
B. Progressive supranuclear gaze palsy
C. Multiple system atrophy
D. Foville syndrome

Response:
Option A - Ruled out as resting tremors are absent & disease is ruled out as it is a ATYPICAL PARKINSONISM disease. Option B- since in the question its mentioned regarding the veical gaze of the patient, it's the most appropriate option. Option C- Ruled out as autonomic insufficiency isn't mentioned in the question. Option D- Manifestation of STROKE related to Pons with VIth nerve palsy.

So the answer to this question is B


Instruction:
Rosenthal fibres are seen in -
A. Pilocytic astrocytoma
B. Glioblastoma
C. Medulloblastoma
D. Ependymoma

Response:
Rosenthal fibers are thick, elongated, "corkscrew" eosinophilic bundles that are found on H&E staining of the brain.
They are seen in the presence of long-standing gliosis, pilocytic astrocytoma, Alexander's disease (a rare leukodystrophy) and fucosidosis.

So the answer to this question is A


Instruction:
Coracobrachialis is pierced by which nerve
A. Axillary
B. Median
C. Musculocutaneous
D. Ulnar

Response:
Coracobrachialis is pierced by the musculocutaneous nerve (C5 - C7). It flexes the arm at the shoulder joint. REF: BD Chaurasia 7th edition Page no: 91.

So the answer to this question is C


Instruction:
Blackfoot disease is caused by: NEET 14
A. Arsenic
B. Cadmium
C. Lead
D. Mercury

Response:
Ans. Arsenic

So the answer to this question is A


Instruction:
Most common cause of perinatal moality in Twins:
A. Dystonia
B. Hemorrhage
C. Anemia
D. Prematurity

Response:
Prematurity

So the answer to this question is D


Instruction:
When a surgeon wants to send the autopsy specimen for virological examination, it should be preserved in:
A. 50% glycerine
B. 10% formalin
C. Rectified spirit
D. Saturated solution of common salt

Response:
A i.e. 50% Glycerine 50% glycerine(2 is used as preservative to send autopsy specimen for virological examination.

So the answer to this question is A


Instruction:
Lepra cells seen in leprosy are:
A. Lymphocytes
B. Plasma cells
C. Vacuolated Histiocytes
D. Neutrophils

Response:
None

So the answer to this question is C