Instruction:
Capnography is useful for:
A. determining Vaporizer malfunction or contamination
B. determining circuit hypoxia
C. determining the appropriate placement of endotracheal tube.
D. detetcing concentration of oxygen in the anesthetic circuit

Response:
Ans. C. Determining the appropriate placement of endotracheal tube. (Ref. Clinical Anesthesia by Paul 5th/pg. 670-671)Clinical Anesthesia by Paul 5th Edition/pg. 671.........."Capnography is an essential element in determining the appropriate placement of endotracheal tubes. The presence of a stable ETCO2 for three successive breaths indicates that the tube is not in the esophagus. A continuous, stable CO2 waveform ensures the presence of alveolar ventilation but does not necessarily indicate that the endotracheal tube is properly positioned in the trachea. For example, the tip of the tube could be located in a main-stem bronchus. Capnography is also a monitor of potential changes in perfusion or dead space, is a very sensitive indicator of anesthetic circuit disconnects and gas circuit leaks, and is a method to detect the quality of CO2 absorption".Endotracheal intubation or laryngeal mask airway insertion requires qualitative identification of carbon dioxide in the expired gas. During general anesthesia, capnography and end-tidal carbon dioxide analysis are encouraged. Capnometry is the measurement and numeric representation of the C02 concentration during inspiration and expiration. A capnogram is a continuous concentration-time display of the C02 concentration sampled at a patient's airway during ventilation.Capnography# is the continuous monitoring of a patient's capnogram.# The capnogram is divided into four distinct phases (Fi. The first phase (A-B) represents the initial stage of expiration. Gas sampled during this phase occupies the anatomic dead space and is normally devoid of CO2. At point B, CO2- containing gas presents itself at the sampling site, and a sharp upstroke (B-C) is seen in the capnogram. The slope of this upstroke is determined by the evenness of ventilation and alveolar emptying. Phase C-D represents the alveolar or expiratory plateau. At this phase of the capnogram, alveolar gas is being sampled. Normally, this part of the waveform is almost horizontal. Point D is the highest CO2 value and is called the end-tidal CO2 (ETCO2). ETCO2 is the best reflection of the alveolar CO2 (PACO2). As the patient begins to inspire, fresh gas is entrained and there is a steep downstroke (D-E) back to baseline. Unless rebreathing of CO2 occurs, the baseline approaches zero.# The utility of capnography depends on an understanding of the relationship between arterial CO2 (PaCO2), alveolar CO2 (PACO2), and ETCO2. If the PaCO2-PACO2 gradient is constant and small, capnography provides a noninvasive, continuous, real-time reflection of ventilation. During general anesthesia, the ETC02-PaC02 gradient typically is 5 to 10 mm Hg.# Maldistribution is a common cause of an increased PaCO2-PACO2 gradient. Other patient factors that may influence the accuracy of ETCO2 monitoring by widening the PaCO2-ETCO2 gradient include shallow tidal breaths, prolongation of the expiratory phase of ventilation, or uneven alveolar emptying.# Increases in ETCO2 can be expected when CO2 production exceeds ventilation, such as in hyperthermia or when an exogenous source of CO2 is present# A sudden drop in ETCO2 to near zero followed by the absence of a CO2 waveform is a potentially life-threatening problem that could indicate malposition of an endotracheal tube into the pharynx or esophagus, sudden severe hypotension, pulmonary embolism, a cardiac arrest, or an artifact resulting from disruption of sampling lines. During life-saving cardiopulmonary resuscitation, the generation of adequate perfusion can be assessed by the restoration of the CO2 waveform.# Whereas abrupt decreases in the ETCO2 are often associated with an altered cardiopulmonary status (e.g., embolism or hypoperfusion), gradual reductions in ETCO2 more often reflect decreases in PaCO2 that occur after increases in minute ventilation where ventilation overmatches CO2 production.# Several methods for the quantification of CO2 have been applied to patient monitoring systems. One of the most commonly used methods is based on infrared absorption spectrophotometry (IRAS).Additional Educational points:# ETCO2 is the best reflection of the alveolar CO2 tension.# Increase in ETCO2 is the most sensitive early sign of malignant hyperthermia.6Most Sensitsive|||||Transesophageal echocardiographyPrecordial DopplerPulmonary artery catheter (| PAP)Capnography (|ETCO2)Mass spectrometry (| ETN2 most specific and quantitative)Least SensitivePAP= pulmonary artery pressure; ETCO2= end-tidal carbon dioxide; ETN2=end-tidal nitrogen.

So the answer to this question is C


Instruction:
Inguinal L.node is found to be involved in a case of ovarian Ca. Revised FIGO staging would be ?
A. Stage III A
B. Stage IV B
C. Stage III C
D. Stage IV A

Response:
Ans- B

So the answer to this question is B


Instruction:
All of the following are false about Tenapanor except:
A. It is indicated for use in diarrhoea predominant IBS
B. Can be used in children < 6 years
C. Acts by inhibiting NHE3 on enterocytes, reducing reabsorption of sodium in small intestine
D. Constipation is a serious adverse effect

Response:
Tenapanor Small molecule medication approved in September 2019 for the treatment of constipation-predominant irritable bowel-syndrome (IBS-C). MOA - Inhibitor of the sodium/hydrogen exchanger isoform 3 (NHE3) transpoer First and currently only medication within its class By inhibiting NHE3 on the apical surface of the enterocytes, tenapanor reduces absorption of sodium from the small intestine and colon, Resulting in an increase in water secretion into the intestinal lumen, which accelerates intestinal transit time and results in a softer stool consistency. Recommended dosage in adults - 50 mg orally twice daily- prior to breakfast or the first meal of the day and immediately prior to dinner C/I Patients< 6 years of age due to the risk of serious dehydration, Patients with known or suspected mechanical gastrointestinal obstruction

So the answer to this question is D


Instruction:
The structure derived from the right fourth aortic arch-
A. Right common carotid artery
B. Right subclavian artery
C. Aortic arch
D. None of the above

Response:
None

So the answer to this question is B


Instruction:
Which of the following statement about Standard Distribution curve is true?
A. Mean = 2 Median
B. Mean = Median
C. Median = Variance
D. Standard Detion = 2 Variance

Response:
The standard distribution curve (Normal distribution) is a perfectly symmetrical, bell shaped curve such that the mean, median and mode, all have the same value and coincide at the centre. Mean is obtained by adding individual observations and then divided by the number of observations. Median is obtained by arranging the data in ascending or descending order of magnitude and then the value of the middle observation is located which is called the median. Mode is the most frequently occurring value. Ref: Park's Textbook of Preventive and Social Medicine, By K. Park, 19th Edition, Pages 699-702

So the answer to this question is B


Instruction:
In a young patient presenting with recurrent vitreous hemorrhage, diagnosis is –
A. Eale's disease
B. CRVO
C. Proliferative retinopathy
D. Coat's disease

Response:
Eale's disease is characterized by recurrent bilateral vitreous hemorrhage in young adult (20-30 yrs) males.

So the answer to this question is A


Instruction:
Elemental iron supplementation in Iron deficiency anemia is –
A. 300 – 400 mg
B. 150 – 200 mg
C. 100 – 150 mg
D. < 100 mg

Response:
None

So the answer to this question is C


Instruction:
Iron source of poor man
A. Soya
B. Orange fruit
C. Jaggery
D. Almond

Response:
None

So the answer to this question is C


Instruction:
Judet view of X-ray is used for viewing injury of what ?
A. Pelvis
B. Calcaneum
C. Scaphoid
D. Spine

Response:
*Judet view of X- ray is for acetabular (pelvis) fracture. Ref : Maheshwari J Essential Ohopaedics 9th/e p.38 & 5th/e p.38,369

So the answer to this question is A


Instruction:
Which of the following is used in the treatment of well differentiated thyroid carcinoma -
A. Il31
B. 99m Tc
C. 32p
D. MIBG

Response:
Ans. is 'a' i.e., I131

So the answer to this question is A


Instruction:
The RPR test uses VDRL antigen containing fine
A. Carbon paicles
B. Iodine paicles
C. Silver ions
D. Sodium ions

Response:
The rapid plasma reagin (RPR) test is a macroscopic, nontreponemal flocculation card test used to screen for syphilis. RPR antigen is prepared from a modified VDR antigen suspension containing choline chloride to eliminate the need to heat inactivate serum, ethylenediaminetetraacetic acid (EDTA) to enhance the stability of the suspension, and finely divided charcoal paicles as a visualizing agent. Ref: Ananthanarayana textbook of Microbiology 9th edition Pgno : 374

So the answer to this question is A


Instruction:
Which of the following is the best marker for diagnosing thyroid related disorder?
A. T3
B. T4
C. TSH
D. Thyroglobulin

Response:
Thyroid Stimulating Hormone assay is highly sensitive and specific in the laboratory assessment of thyroid function. TSH levels change dynamically in response to alterations of T4 and T3, so while evaluating a case of thyroid dysfunction, first we have to determine whether TSH is suppressed, normal or elevated. With rare exceptions a normal TSH level excludes a primary abnormality of thyroid function. The finding of an abnormal TSH level must be followed by measurements of circulating thyroid hormone levels to confirm the diagnosis of hypehyroidism or hypothyroidism. TSH should not be used as an isolated laboratory test to assess thyroid function in patients with suspected or known pituitary disease. Ref: Harrison's Internal Medicine, 18th Edition, Pages 2916-2917, Chapter 341 ; Principles of Critical Care, 3rd Eedition, Jesse B. Hall, Chapter 80.

So the answer to this question is C


Instruction:
Which of the following medications is typically avoided in patients with pheochromocytoma due to the risk of hypeensive crisis?
A. Promethazine
B. Metoclopramide
C. Haloperidol
D. Ondansetron

Response:
For pheochromocytoma patients, METOCLOPRAMIDE is C/I as it can lead to Hypeensive crisis. The gastroprokinetic agent metoclopramide is known to stimulate catecholamine secretion from pheochromocytomas. Metoclopramide is contraindicated in patients with pheochromocytoma due to the risk of hypeensive crisis. Metoclopramide induced hypeensive crisis can be treated with phentolamine. If the patient has emesis, a serotonin 5-HT3 receptor antagonist should be used. Ondansetron does not have any effect on dopamine or norepinephrine

So the answer to this question is B


Instruction:
NRHM (National Rural Health Mission) done in
A. 2002
B. 2005
C. 2006
D. 2011

Response:
None

So the answer to this question is B


Instruction:
Large kidney is seen in A/E -
A. Benign nephrosclerosis
B. Lymphoma
C. Amyloidosis
D. DM

Response:
Ans. is 'a' i.e., Benign nephrosclerosis 2Kidney sized is increased ino Diabetes mellitus (Diabetic nephropathy) o Amyloidosiso Polycystic kidney disease o Lymphoma

So the answer to this question is A


Instruction:
Hernia through foramen of Bochdalek true is -
A. Congeniatal hernia
B. Is asymptomatic
C. Seen especially in males
D. Least common

Response:
Answer is 'a' i.e. Congenital hernia

So the answer to this question is A


Instruction:
For diagnosis of lunacy a person can be kept under observation at a streach for :
A. 2 Days
B. 7 Days
C. 10 Days
D. 30 Days

Response:
C i.e. 10 days

So the answer to this question is C


Instruction:
Hockey stick appearance on echo is a feature of
A. Mitral stenosis
B. Mitral incompetence
C. Aoic stenosis
D. Aoic regurgitation

Response:
Echocardiographic features of mitral stenosis:-Thickened and calcified mitral leaflets and subvalvular apparatus -Decreased E-F slope (M-mode)-Hockey stick appearance of the anterior mitral leaflet in diastole(long axis view)-Immobility of the posterior mitral leaflet (a similar appearance can be seen in hypereosinophilia or ergot use)-Fish mouth orifice in the sho axis view-Increased LA size, with the potential for thrombus formationDynamic CT: the Restricted opening of the thickened valve from commissural fusion (especially with rheumatic valve disease), valve calcification, or both results in a "fish-mouth" appearance on sho-axis images. Bowing of a thickened and fibrotic anterior leaflet during diastole may result in a "hockey-stick" appearance which is best seen on two- or four-chamber images(Ref: The echo manual by Jae K Oh; J B Seward; A Jamil Tajik, page no 202)

So the answer to this question is A


Instruction:
Which of the following is characterised by typical ash - grey slough
A. Furuncle
B. Carbuncle
C. Erysipeals
D. Cellulitis

Response:
Ash - grey slough is typical of carbuncle.

So the answer to this question is B


Instruction:
Ground glass appearance is not seen in –
A. Hyaline membrane disease
B. Left to right shunts
C. Pneumonia
D. Obstructive TAPVC

Response:
Ground grass appearance is seen in right to left shunt (not in left to right shunt).
All other options, pneumonia (interstitial pneumonia), HMD, and obstructive TAPV, can cause ground glass appearance.

So the answer to this question is B


Instruction:
Decrease number of melanocyte is seen in:
A. Albinism
B. Piebaldism
C. Ichthyosis
D. Chemical leucoderma

Response:
Piebaldism is a Autosomal dominant disease characterized by patches of skin that lack melanin as a result of defect in migration of melanoblast from neural crest to skin.
Skin biopsy: Hypo melanotic area-few to no melanocytes. Vitiligo: is an acquired disease and similar patch of loss ofmelanin, but the loss develops after birth and progressive. Albinism: Congenital inability to synthesize melanin.

So the answer to this question is B


Instruction:
Which of the following abnormality is seen in a preoperative pulmonary function test, done in a patient with severe kyphoscoliosis?
A. Increased FRC
B. Increased RV/TLC
C. Reduced FEV1/FVC
D. Reduced FEV25-75

Response:
Kyphoscoliosis is a combination of excessive anteroposterior and lateral curvature of thoracic spine. The major pathophysiologic effect of severe kyphoscoliosis are restrictive lung disease and ventillation perfusion imbalance resulting in chronic hypoventilation, hypoxic vasoconstriction pulmonary aery hypeension and cor pulmonale. Pulmonary function test shows a decrease in vital capacity and total lung capacity and an increase in residual volume, resulting in an increased residual volume/total lung capacity. Flow is reduced only in propotion to the reduction in vital capacity. Ref: Pulmonary Function Testing and Cardiopulmonary Stress Testing By Vincent C. Madama, 2nd Edition, Page 102; Chest Medicine: Essentials of Pulmonary and Critical Care Medicine By Ronald B. George, 5th Edition, Pages 440-1

So the answer to this question is B


Instruction:
Floatation of a body in water is due to :
A. Flesh being eaten by fishes
B. Presence of gases
C. Water logging of body
D. All of the above

Response:
B i.e. Presence of gasses

So the answer to this question is B


Instruction:
Rods and cones differ in all, Except:
A. Signal transduction
B. Light sensitivity
C. Wavelength
D. Acuity

Response:
Ans is 'a' i.e. signal transduction "The process of phototransduction in cones is virtually the same as in rods: the only major difference is in the types of opsin in the membranous disks of the cone outer segments." -Neuroscience By Mark F. Bear, Barry W. Connors,Michael A. Paradiso 3/e p296The retina is organized in 10 layers and contains rods and cones, the photoreceptor cells along with 4 other types of neurons- bipolar cells, ganglion cells, horizontal cells and amacrine cells.The rods and cones synapse with bipolar cells, and the bipolar cells synapse with ganglion cells. The axons of ganglion cells converge and leave the eye as the optic nerve.Differences between rods and conesThe two photoreceptors differ from each other in several ways. Rods and cones differ in structure, photochemical molecules, sensitivity, retinal distribution, synaptic connections, and function.The total number of rods in the human retina (approx. 90 million) far exceeds the number of cones (roughly 4.5 million). As a result, the density of rods is much greater than cones throughout most of the retina. However, this relationship changes dramatically in the fovea, a highly specialized region of the central retina that measures about 1.2 millimeters in diameter. In the fovea, cone density increases almost 200-fold, reaching, at its center, the highest receptor packing density anywhere in the retina. The increased density of cones in the fovea is accompanied by a sharp decline in the density of rods. In fact, the central 300 pm of the fovea, called the foveola, is totally rod-free. The extremely high density of cone receptors in the fovea, and the one-to- one relationship with bipolar cells and retinal ganglion cells (see later), endows this region (and the cone system generally) with the capacity to mediate high visual acuity. As cone density declines with eccentricity and the degree of convergence onto retinal ganglion cells increases, acuity is markedly reducedThe two types of receptors differ in their size and their disk morphology. The outer segments of rods contain approximately 1000 to 2000 identical disks. The disks in rods contain more photopigment (rhodopsin) and consequently capture more photons of light than cones.There is also a higher amplification in the rods than in the cones. This means that, in rods, more cGMP gets hydrolyzed for each absorbed photon of light. This amplification leads to a longer integration and a more sensitive cell. The high amplification and long integration leads to high sensitivity to scattered light. The high sensitivity is the reason that rods are effective in weak light. The rods are thus responsible for vision when the light is dim (scotopic vision), whereas the cones are responsible in good light (photopic vision).In contrast, the cones do not have individual disks to hold the photochemical used in the transduction process. The photochemical in cones are in the outer membrane of the cell and the shape of the cone is determined by the folding of the outer membrane. The cones require more light because they have fewer photochemical and less amplification.The opsin molecules in the photochemical differ in the two receptors. All rods have the same opsin molecule. Because the opsin molecule determines the particular wavelength that can be absorbed most readily by the retinal, all the rods are similar in their response to light. The rods have an absorption peak near 500 nm.Each cone, on the other hand, has one of three different opsin molecules. Thus there are three different types of cones- red cone, blue cone and the green cone. The wavelength absorption characteristic of each cone depends on which opsin is bound to the retinal. The opsin determines the cone-wavelength absorption sensitivity. This means that cones differ from each other as well as from the rods in their absorption characteristics. Thus the cones are responsible for color vision.Other important difference between rods and cones concern their connections with other neurons in the retina.Properties of Rod and Cone SystemsRodsConesCommentMore photopigmentLess photopigment Slow response: long integration timeFast response: short integration time High amplificationLess amplification Saturating Response (by 6% bleached)Non-saturating response (except S- cones)The rods' response saturates when only a small amount of the pigment is bleached (the absorption of a photon by a pigment molecule is known as bleaching the pigment).Night visionDay vision Highly convergent retinal pathwaysLess convergent retinal pathwaysSpatial integrationHigh sensitivityLower absolute sensitivity Low acuityHigh acuityResults from degree of spatial integrationAchromatic: one type of pigmentChromatic: three types of pigmentColor vision results from comparisons between cone responsesNotably, many rods are connected to one bipolar cell- i.e. there is a high degree of convergence. For the cones, on the other hand, there is typically much less convergence, with a few cones connected to one bipolar cell. This explains more sensitivity of the rods over cones. The difference in convergence means that the cones provide information with a higher spatial resolution than the rods. Signal transduction in the RetinaThe mechanism of signal transduction or phototransduction i.e. the process by which light is converted into electrical signals, is almost similar in both the rods and cones. But both the photoreceptors have differences with other sensory receptors.The photoreceptors do not behave like other receptors when exposed to their stimulus: they are hyperpolarized instead of depolarizedThe photoreceptors are in a depolarized state in the dark. This is probably caused by Na+ channels that are open under such conditions. Unlike other receptors, the photoreceptors (and the bipolar cells) do not produce action potentials but only graded changes of the membrane potential (k/a local graded potential). Because the distance is very short from the outer segment-where the membrane potential changes are produced-to the synapses between the photoreceptors and the bipolar cells, even small fluctuations of the membrane potential cause alterations of the transmitter release from the photoreceptors. There is thus no need for the production of action potentials, which are important when signals are to be propagated over long distances.Light falling on the photoreceptor cells alters the configuration of the retinene (from 11-cis configuration to 11- Trans' isomer). This in turn alters the configuration of the opsin, and the opsin change activates the associated G protein. This ultimately leads to closure of the Na+ channels.Closure of the Na+ channels hyperpolarizes the cell. The hyperpolarization reduces the release of synaptic transmitter, and this generates a signal in the bipolar cells that ultimately leads to action potentials in ganglion cells.* In contrast to the photoreceptors and the bipolar, the ganglion cells produce action potentials (conducted in the optic nerve to the higher visual centers).

So the answer to this question is A


Instruction:
Mechanism of action of streptomycin-
A. Inhibition of translation
B. Inhibition of signal transduction
C. Inhibition of citric acid cycle
D. Inhibits mitochondria

Response:
Ans. is 'a' i.e., Inhibition of translation o Streptomycin inhibit protein synthesis (translation).

So the answer to this question is A


Instruction:
Growth on a cell-free aificial solid medium is possible for following except -
A. Ureaplasma urealyticum
B. Mycoplasma Pneumoniae
C. C & L form of proteus vulgaris
D. Chlamydia

Response:
They are bacteria but may be considered as viruses on account of their filterability and failure to grow in cell-free culture media. REF: Textbook of microbiology Ananthanarayan, and Paniker. Page no.: 415

So the answer to this question is D


Instruction:
Most common type of intususucption is?
A. Ileo-ileal
B. Ileo colic
C. Ileo-ileo-colic
D. Colo-colic

Response:
ANSWER: (B) Ileo colicREF: Bailey and love 24th ed p. 1195Most common type of intussuseption is ileocolic

So the answer to this question is B


Instruction:
Telmisartan lowers blood pressure by:
A. Inhibiting formation of angiotensin I to angiotensin II
B. Inhibiting conversion of renin to angiotensin I
C. Blocking AT1 receptors
D. Interfering with degradation of bradykinin

Response:
None

So the answer to this question is C


Instruction:
Which statement best describes the stomach?
A. Lymph from the superior 2/3 of the stomach drains into the suprapancreatic nodes
B. All lymph from the stomach drains through the coeliac nodes
C. The gastric branches of the vagi are given of in the greater curve
D. The lower oesophageal sphincter is supplied by the nerves of Latarjet

Response:
All the lymph drainage from the stomach eventually drains into the coeliac nodes and then into the cisterna chyli. All the lymph drainage from the stomach initially drains in three directions. On the right side the lymph drains to the left and right gastric nodes. On the left the superior poion drains into the pancreaticosplenic nodes and suprapancreatic nodes and the inferior poion drains into nodes around the pylorus. The anterior and posterior vagi pass along the lesser curvature of the stomach and give off gastric branches as they descend. Ref: Gray's Anatomy 41st edition Pgno: 1118

So the answer to this question is B


Instruction:
True about pKa is:
A. pH at which ionized fraction of drug equals to unionized fraction
B. pH at which ionized fraction of drug is more than unionized fraction
C. pH at which ionized fraction of drug is less than unionized fraction
D. pH at which ionized fraction of drug is twice unionized fraction

Response:
Ans. a. pH at which ionized fraction of drug equals to unionized fractionRef: Goodman Gilman 12th E/P 18pKa is the pH at which as drug is 50% ionized and 50% unionized, i.e. the fraction of ionized drug equals to the fraction of unionized drug.The effect of pH on drug ionization and unionization can be calculated for acidic and basic drug by Hasselbalch equations given below:Basic drug: Log10 (Unionizeddrug)(Ionizeddrug) =pK4 -PHAcidic drug: Log10 (Ionizeddrug)(Ionizeddrug) =pKa -PHWhat is pKa ?If the ionized drug=unionized drug, i.e. 50% each, then the equation becomes,Log10 1=PKa -PHOr,0=pKa -pHOr. pKa =pHHencepKa is the pH at which the drug is 50% ionized and 50% unionized.

So the answer to this question is A


Instruction:
Monogenic &Autosomal II> ominant cause of HTN is/are -a) 17-α Hydroxylase deficiencyb) Familial Conn's syndromec) Pregnancy induced HTNd) Glucocorticoid remediable hyperaldosteronism
A. cd
B. bc
C. bd
D. ac

Response:
None

So the answer to this question is A


Instruction:
Best method of estimation of amount of proteinuria in a 2 year old child with nephrotic syndrome is
A. Dipstick test
B. 24 hr urine protein
C. Microalbuminuria
D. Spot urine sample for Protein/ Creatinine ratio

Response:
None

So the answer to this question is D


Instruction:
Sacrotuberous ligament is pierced by-
A. S1 nerve
B. L1 nerve
C. Coccygeal nerve
D. None

Response:
Ans. C. Coccygeal nerveSacrotuberous ligamenta. The sacrotuberous ligament is attached by its broad base to the posterior superior iliac spine., the posterior sacroiliac ligaments (with which it is partly blended), to the lower transverse sacral tubercles and the lateral margins of the lower sacrum and upper coccyx.b. Its oblique fibres descend laterally, converging to form a thick, narrow band that widens again below and is attached to the medial margin of the ischial tuberosity. It then spreads along the ischial ramus as the falciform process, whose concave edge blends with the fascial sheath of the internal pudendal vessels and pudendal nerve.c. The lowest fibres of gluteus maximus are attached to the posterior surface of the ligament; superficial fibres of the lower part of the ligament continue into the tendon of biceps femoris.d. The ligament is pierced by the coccygeal branches of the inferior gluteal artery, the perforating cutaneous nerve and filaments of the coccygeal plexus.

So the answer to this question is C


Instruction:
Thoracic outlet syndrome is seen most commonly in which age group
A. Children
B. Elderly
C. Middle aged women
D. Atheletes

Response:
.THORACIC OUTLET SYNDROME (TOS) It is syndrome complex due to neurovascular bundle compression in the thoracic outlet. Thoracic outlet has got two main spaces: * Scalene triangle is bound by scalenus anterior, scalenus medius and first rib. It contains subclan aery and brachial plexus. * Costoclavicular space is bound by clavicle, first rib, costoclavicular ligament and scalenus medius. It contains subclan aery and vein and brachial plexus. Causes - * Cervical rib. * Long C7 transverse process. * Anomalous inseion of scalene muscles. * Scalene muscle hyperophy. * Scalene minimus. * Abnormal bands and ligaments. * Fracture clavicle or first rib. * Exostosis. * Tumours in the region. * Brachial plexus trauma and diseases. ref:SRB&;s manual of surgery,ed 3,pg no 378

So the answer to this question is C


Instruction:
Investigation of choice in clinical suspicion of stroke
A. Noncontrast head CT
B. Contrast head CT
C. Noncontrast head MRI
D. Contrast head MRI

Response:
After the clinical diagnosis of stroke is made, an orderly process of evaluation and treatment should follow. The first goal is to prevent or reverse brain injury. Attend to the patient's airway, breathing, and circulation (ABCs), and Treat hypoglycemia or hyperglycemia if identified. Perform an emergency noncontrast head CT scan to differentiate between ischemic stroke and hemorrhagic strokeRef: Harrison 19e pg: 2561

So the answer to this question is A


Instruction:
A CKD patient had to undergo dialysis. His Hb was 5.2gm% so two blood transfusions were to be given. First bag was completed in 2 hours. Second was staed and midway he developed shoness of breath, hypeension. Vitals: BP 180/120 mm Hg and pulse rate 110/min. What is the cause?
A. Allergic
B. Transfusion related circulatory overload (TACO)
C. TRALI
D. FNHTR

Response:
This C.K.D patient has developed severe anemia which is a compensated hea failure state. The patient was given first unit of blood in two hours instead of standard four hours and he is already in compensated hea failure. The acute decompensation due to volume overloading that has subsequently occurred will result in pulmonary edema and shoness of breath. This presentation is called as transfusion associated circulatory overload. Ideally patients of CKD should be given erythropoietin injections to reduce the incidence of having severe anemia.

So the answer to this question is B


Instruction:
Rabies is identified by -
A. Guarnieri bodies
B. Negri bodies
C. Cowdry A bodies
D. Cowdry B bodies

Response:
Few Important Bodies in Microbiology

 Bollinger bodies — Fowlpox
 Brassy body — Dark shrunken blood corpuscle found in malaria
 Call exner bodies — Granulosa theca cell tumour
 Chromatid bodies — Entamoeba histolytica precyst
 Citron bodies — cl. Septicum
 Cigar bodies — Sporotrichosis
 Councilman bodies — Yellow fever/viral hepatitis
 Coccoid X bodies — Psittacosis
 Donovan bodies — Granuloma inguinale (LGV)
 Guarnieri bodies — Inclusion bodies of vaccinia
 Henderson paterson bodies — Molluscum contagiosum
 Levinthal cole lillie bodies — Psittacosis
 Mooser bodies — Endemic typhus
 Negri bodies — Rabies
 Paschen bodies — Vaccinia, variola
 Ross’s bodies — Syphilis
 Sclerotic bodies — Chromoblastomycosis

So the answer to this question is B


Instruction:
A hypertensive, diabetic is having proteinuria,antihypertensive of choice is -
A. Propanolol
B. Clonidine
C. Enalapril
D. Alpha methyldopa

Response:
None

So the answer to this question is C


Instruction:
All the following conditions cause thrombocytopenia except-
A. Giant hemangioma
B. Infectious mononucleosis
C. HIV infection
D. Iron deficiency anemia

Response:
Iron deficiency anemia develops when body stores of iron drop too low to suppo normal red blood cell (RBC) production. Inadequate dietary iron, impaired iron absorption, bleeding, or loss of body iron in the urine may be the cause. Iron equilibrium in the body normally is regulated carefully to ensure that sufficient iron is absorbed in order to compensate for body losses of iron (see the image below). The total body iron in a 70-kg man is about 4 g. This is maintained by a balance between absorption and body losses. Although the body only absorbs 1 mg daily to maintain equilibrium, the internal requirement for iron is greater (20-25 mg). An erythrocyte has a lifespan of 120 days so that 0.8% of red blood cells are destroyed and replaced each day. A man with 5 L of blood volume has 2.5 g of iron incorporated into the hemoglobin, with a daily turnover of 20 mg for hemoglobin synthesis and degradation and another 5 mg for other requirements. Most of this iron passes through the plasma for reutilization. Iron in excess of these requirements is deposited in body stores as ferritin or hemosiderin Ref Robbins 9/e pg 440

So the answer to this question is D


Instruction:
Pseudo Koebner's phenomenon is seen in -
A. DLE
B. Lichen planus
C. Kaposis sarcoma
D. Plane was

Response:
Ans. is'd' i.e., Plane was

So the answer to this question is D


Instruction:
Eruption cyst
A. Transforms into dentigerous cyst
B. Regresses after eruption of the tooth
C. Is found in the place of the missing tooth
D. Is a type of dentigerous cyst

Response:
None

So the answer to this question is B


Instruction:
A 71/2 year old girl with non productive cough, mild stridor for 3 months duration, patient is improving but suddenly developed wheeze, productive cough, mild fever and on x-ray hyperlucency and PFT shows obstructive curve. The most probable diagnosis is,
A. Bronchiolitis obliterans
B. Hemosiderosis
C. Pulmonary alveolar microlithiasis
D. Follicular bronchitis

Response:
Bronchiolitis obliterans Repeat from Nov 08 Q no.130

So the answer to this question is A


Instruction:
All of the foramens are found in sphenoid bone except:
A. Foramen rotundum
B. Foramen lacerum
C. Optic canal
D. Foramen ovale

Response:
None

So the answer to this question is B


Instruction:
IV dose of 1: 10000 concentration of epinephrine in pre term baby is?
A. 0.1m1
B. 0.2 ml
C. 0.3 ml
D. 0.4 ml

Response:
Ans. is 'b' i.e., 0-2 ml Epinephrine The IV dose of 1: 10,000 concentration is 0.1m1/Kg. Thus it is about 0.5ml for term baby and 0.2 ml for pre term baby. Endotracheal tube dosing is 1m1/ Kg. thus it is about 3m1 for term baby and 1 ml for preterm.

So the answer to this question is B


Instruction:
Physiological unlocking is caused by -
A. Popliteus
B. Rectus femoris
C. Semimembranosus
D. Sartorius

Response:
During last stages of extension, femur rotates medially on tibia due to quadriceps femoris to lock the knee joint. Femur rotates laterally on tibia during flexion.
Quadriceps femoris- Vastus medialis, vastus lateralis, vastus intermedius, rectus femoris. Unlocking occurs with popliteus.

So the answer to this question is A


Instruction:
Trypsin, Chymotrypsin & Elastases - What type of enzymes are they?
A. Hydrolases
B. Lyases
C. Synthases
D. Synthetases

Response:
Ans. is 'a' i.e., Hydrolases* Trypsin, Chymotrypsin and Elastase are endopeptidases secreted by Pancreas* Trypsin acts at Lys - X or Arg - X linkage* Chymotrypsin acts at Aro - X linkage* Elastase acts on Ali - X linkage* All these break a peptide linkage by adding water. Hence they are hydrolases

So the answer to this question is A


Instruction:
Which of the following is not the function of zinc in 2nd  generation amalgam alloy?
A. Increase plasticity
B. Decrease brittleness
C. Act as deoxidizer
D. Act as scavenger

Response:
Traces of zinc (up to 1%) were added to act as a deoxidizer or scavenger for the alloy and to decrease its brittleness.
Reference: Marzouk Operative dentistry, pg-106

So the answer to this question is A


Instruction:
Boerhaave&;s syndrome is due to
A. Drug induced esophagus perforation
B. Corrosive injury
C. Spontaneous perforation
D. Gastro-esophageal reflux disease

Response:
BOERHAAVE&;S syndrome is a tear in the lower third of oesophagus which occurs when a person vomits against a closed glottis, causing leaks into the mediastinum, pleural cavity, and peritoneum.site of perforation is about 2-10 cm of posterolateral pa of the lower oesophagus.investigations include chest Xray, MRI/CT, and total count.treatment includes feeding by jejunostomy, surgery with resection. Ref: SRB&;s manual of surgery,5th ed, pg no 802

So the answer to this question is C


Instruction:
A male aged 60 years has foul breath. He regurgitates food that is eaten 3 days ago : Likely diagnosis is ?
A. Zenkers diveiculum
B. Meckels diveiculum
C. Scleroderma
D. Achalasia cardia

Response:
Ans. is 'a' i.e., Zenker's diveiculum

So the answer to this question is A


Instruction:
In an after coming head the following bone Is perforated during craniotomy:
A. Occiput
B. Parietal
C. Palate
D. Frontal

Response:
Craniotomy : Consists of perforating the fetal skull, evacuating the contents and then delivering rest of the fetal pas. Indications : Delivery of dead fetus in obstructed labour with a cephalic presentation Delivery of the arrested after coming head in vaginal breech delivery Hydrocephalus Ref: Dutta Obs 9e pg 544.

So the answer to this question is A


Instruction:
In Intrapaum Fetal Monitoring of women with pregnancies at risk, fetal hea auscultation is performed at least every --- minutes during first-stage labor and every --- minutes during the second stage
A. 60min and 30min
B. 30min and 15min
C. 15min and 5min
D. 5min and 1min

Response:
Intrapaum fetal monitoring:Briefly, the American Academy of Pediatrics and American College of Obstetricians and Gynecologists (2012) recommend that during first-stage labor, in the absence of any abnormalities, the fetal hea rate should be checked immediately after a contraction at least every 30 minutes and then every 15 minutes during the second stage. If continuous electronic monitoring is used, the tracing is evaluated at least every 30 minutes during the first stage and at least every 15 minutes during second-stage labor. For women with pregnancies at risk, fetal hea auscultation is performed at least every 15 minutes during first-stage labor and every 5 minutes during the second stage. Continuous electronic monitoring may be used with evaluation of the tracing every 15 minutes during the first stage of labor, and every 5 minutes during the second stage.(Ref: William's Obstetrics; 25th edition)

So the answer to this question is C


Instruction:
Posterior staphyloma is associated with
A. Pathological myopia
B. Uveoscleritis
C. Pseudocornea
D. Angle closure glaucoma

Response:
Posterior staphyloma is a condition that can be present in highly myopic individuals with long axial lengths. It is caused by elongation of the globe due to thinning and bulging of the sclera and results in an irregular configuration of the retina at the posterior pole. ref img

So the answer to this question is A


Instruction:
Alderman's nerve is a branch of -
A. Superior cervical ganglion
B. Inferior cervical ganglion
C. Auricular branch of vagus nerve
D. None

Response:
C i.e. Auricualr branch of vagus nerve The auricular branch of vagus nerve is often termed as Alderman's nerve or Arnold's nerveQSince the Alderman of the Anglo-saxon empire was habitually eating and then putting some cold water in the ear to stimulate this nerve to initiate vomiting, this name came into being. In this way they could empty their stomach quickly and resta eating.

So the answer to this question is C


Instruction:
Corneal layer which does not regenerate
A. Bowman's layer
B. Stroma
C. Descemet's membrane
D. Epithelium

Response:
The layers of cornea are -(anterior to posterior) Epithelium Bowman's membrane Substantia propria(corneal stroma) Pre- Descement's membrane(Dua's layer) Descement's membrane Endothelium Bowman's membrane consists of acellular mass of condensed collagen fibrils. It is not a true elastic membrane but simply considered superficial pa of stroma. It shows considerable resistance to infection. But once destroyed, it is unable to regenerate and, therefore, heals by scaring. Ref;A.K.Khurana;6 th Edition; Page no: 96

So the answer to this question is A


Instruction:
Broca's area is located in
A. Inferior frontal lobe
B. Parietal lobe
C. Superior temporal gyrus
D. Angular gyrus

Response:
None

So the answer to this question is A


Instruction:
A coronally positioned pedicle graft is used when a single tooth exhibits gingival recession and
A. Bleeding
B. Pus discharge
C. Irreversible pulpitis
D. Sensitivity

Response:
None

So the answer to this question is D


Instruction:
Life time risk of TB in HIV Patients -
A. 20%
B. 50%
C. 10%
D. 5%

Response:
Tuberculosis Preventing exposure to tuberculosis is impoant when there is an infectious case in the household, in clinics and in hospitals.Adequate ventilation, masks and safe coughing procedures reduce the risk of exposure. For HIV-uninfected individuals with latent TB infection (LTBI), the lifetime risk of developing active TB due to reactivation is 8 to 10%. In contrast, this risk is approximately 10% per year for HIV-infected persons (128, 142, 227, 228). Ref Harrison 20th edition pg 879

So the answer to this question is A


Instruction:
Vitamin C is required for ?
A. Posttranslational modification
B. Synthesis of epinephrine
C. Tyrosine metabolism
D. All of the above

Response:
Vitamin C (ascorbic acid) Ascorbic acid (Vitamin C) is also called antiscorbutic factor. It is very heat labile, especially in basic medium. Ascorbic acid itself is an active form. Maximum amount of vitamin C is found in adrenal coex. Ascorbic acid functions as a reducing agent and scavanger of free radicals (antioxidant). Its major functions are :- i) In collagen synthesis :- Vitamin C is required for post-translational modification by hydroxylation of proline and lysine residues conveing them into hydroxyproline and hydroxylysine. Thus vitamin C is essential for the conversion of procollagen to collagen, which is rich in hydroxyproline and hydroxylysine. Through collagen synthesis, it plays a role in formation of matrix of bone, cailage, dentine and connective tissue. ii) Synthesis of norepinephrine from dopamine by dopamine-(3-monoxygenase (dopamine-13-hydroxylase) requires Vitamin C. iii) Carnitine synthesis iv) Bile acid synthesis :- 7-a-hydroxylase requires vitamin C. v) Absorption of iron is stimulated by ascorbic acid by conversion of ferric to ferrous ions. vi) During adrenal steroid synthesis, ascorbic acid is required during hydroxylation reactions. vii) Tyrosine metabolism - Oxidation of P-hydroxy-phenylpyruvate to homogentisate. viii) Folate metabolism - Folic acid is conveed to its active form tetrahydrofolate by help of Vitamin C.

So the answer to this question is D


Instruction:
40-year-old female, chronic smoker, presented to the surgical depament with a reddish, painful subareolar mass in the right breast. She is most likely suffering from?
A. Carcinoma breast
B. Periductal mastitis
C. Fat necrosis
D. Granulomatous mastitis

Response:
She is most likely suffering from Periductal mastitis. Aka Zuska's disease /Recurrent subareolar abscess/Squamous metaplasia of lactiferous ducts Smoking has been implicated as a risk factor. Smokers - relative Vit A deficiency - Squamous metaplasia - Obstruction - Dilation of duct Features: Subaerolar mass Pain and redness Nipple retraction Fistula formation May mimic malignancy

So the answer to this question is B


Instruction:
Polycystic ovarian disease not related to
A. Hypertension
B. DM
C. Obesity
D. Hirsutism

Response:
(A) (Hypertension) (459,469, Dutta Gynae 6th)POLYCYSTIC OVARIAN SYNDROME (PCOS)Multifactorial and polygenic conditionsMost obese woman with oligomenorrhea have the polycystic ovarian syndrome (PCOS), with its associated anovulation and ovarian hypcrandrogenism 40% of women with PCOS are absess (Not always). Most non obese with PCOS are also insulin resistance, hyperinsulinemia or the combination of the two are causative or contribute to the ovarian pathophysiology in PCOS in both obese and lean individuals (476-H17th)PCOS (Novak's Gynaecology)Clinical characteristicDiagnostic Criteria* Hirsutism* Obesity* Oligomenorrhea or AmenorrheaMajor criteria Minor criteria* Chronic anovulation* Pre menarchial onset of* Infertility* Hyper androgcncmia (in absence of any notable cause)- Hirsutism- Obesity* Insulin resistance (Abnormal glucose tolerance)* Elevated LH (LH/FSH ratio>2)* Intermittent anovulation associated with hyperandrogenemia (| ed free estosterone, |ed DHEAS)* Multiple cysts are located peripherally along the surface of the ovary giving a ''Necklace" appearance in UCG is diagnostic* The increased conversion of androstenedione to estrogen which occurs to a greater degree in women with lower body obesity may contribute to the increased incidence of uterine cancer in post menopausal women with obesity (467-HI 7th)* PCOS is most strongly associated with a risk of endometrial carcinoma and may also be associateed with an increased risk of ovarian and breast cancer**** Long term sequelae of PCOS- Diabetes mellitus- CVS disorders- MI- Lipidaemia- Hypertension- Endometrial cancer- Breast cancer- Premature ovarian failure following surgery* Autosomal dominant inherited factor** were added as aetiological factor in the development of PCOS CYP21 gene mutation has been discovered in this connectionSerum level in PCOSAndrogens (|)Estrogens (|)LH/FSHInsulinAndroeen excessEstroeen excessLH levels are |edInsulin |ed* Excess production of androgens is seen from both the ovaries and adrenals, (postulated to be due to abnormal regulation of androgen forming enzyme)* Estrogen levels are increased due to peripheral aromatization of androgens* EstrogenQ levels are raised(Estrone>Estradiol)* Enhanced estrogen levels exert a- Positive feedback on LH levels (|ed)- Negative feedback on FSH levels (|ed)* | ed insulin resistance LH/FSH Ratio is characteristically >2

So the answer to this question is A


Instruction:
NADPH+ is generated in the reaction catalysed by
A. LDH
B. G–6 PD
C. G–3 PD
D. Alcohol dehydrogenase

Response:
Amongst the given options, only G6-PD (glucose-6-phosphate dehydrogenase) is NADP linked enzyme.
NAD+ -linked dehydrogenases: Pyruvate dehydrogenase, isocitrate dehydrogenase, malate dehydrogenase. α -ketoglutarate dehydrogenase, glutamate dehydrogenase, glyceraldehyde-3-P dehydrogenase, lactate dehydrogenase, β -hydroxy acyl CoA dehydrogenase, glycerol 3-P dehydrogenase (cytoplasmic).

NADP+-linked dehydrogenases :Glucose-6-P dehydrogenase, 6-Phosphogluconate dehydrogenase, 3-ketoacyl reductase, Enoyl reductase.

FAD-linked dehydrogenases: Succinate dehydrogenase, fatty acyl CoA dehydrogenase, glycerol-3P dehydrogenase (mitochondrial).

So the answer to this question is B


Instruction:
"Mellanby effect" is seen with
A. Paracetamol
B. Aspirin
C. Alcohol
D. LSD

Response:
Mellanby effect: Intoxicating effect of alcohol are greater at a given BAC (blood alcohol concentration) when blood alcohol level is increasing than for the same BAC when blood alcohol level is falling.

So the answer to this question is C


Instruction:
Most common cause of trichiasis is-
A. Stye
B. Trachoma
C. Blephritis
D. Congenital

Response:
*The Most common among these are trachoma, Steven-Johnson syndrome, and trauma Ref: Textbook of Clinical ophthalmology 21st/e p.319

So the answer to this question is B


Instruction:
Maximum unsaturated fatty acids are seen in
A. Mustard oil
B. Groundnut oil
C. Safflower oil
D. Coconut oil

Response:
FatsPUFASafflower oil75Sunflower seed oil65Soya bean oil62Margarine 50Groundnut oil31Palm oil10Butter 3Coconut oil2(Refer: K. Park's Textbook of Preventive and Social medicine, 24thedition, pg no: 649)

So the answer to this question is C


Instruction:
A 75-year-old male develops a cough that produces blood-tinged sputum. He has a fever of 104degF. Gram-positive cocci in clusters are found in a sputum smear. A chest x-ray shows increased density in the right upper lobe. Of the following penicillins, which is most likely to be ineffective?
A. Oxacillin
B. Cloxacillin
C. Ticarcillin
D. Nafcillin

Response:
Ticarcillin resembles carbenicillin and has a high degree of potency against Pseudomonas and Proteusorganisms but is broken down by penicillinase produced by various bacteria, including most staphylococci. Oxacillin, cloxacillin, nafcillin, and dicloxacillin are all resistant to penicillinase and are effective against staphylococci.

So the answer to this question is C


Instruction:
All are used in treating spasmodic dysmenorhoea except :
A. Bromocriptine
B. Ibuprofen
C. Mefanamic acid
D. Norethisterone and Ethinyl estradiol

Response:
Bromocriptine

So the answer to this question is A


Instruction:
All of the following are causes of pneumoperitoneum except -
A. Perforated peptic ulcer
B. Laproscopic Procedure
C. Perforated Appendix
D. Hirschsprung's Disease

Response:
Ans. is 'd' i.e., Hirschsprung's Disease o Gas under left hemidiaphragm is due to fundal gas, a normal finding. When gas is present under right hemidiaphragm also it's due to pneumoperitoneum.Causes of Pneumoperitoneum1. Perforation of GI tract# Peptic ulcer# Inflamamtion (diverticulitis, appendicitis, toxic megacolon, necrotizing enterocolitis)# Infarction# Malignant neoplasm# Obstruction# Pneumatosis cystoides rupture# Iatrogenic-endoscopy2. Iatrogenic# Surgery, peritoneal dialysis, drainage catheters, biopsies3. Penetrating abdominal injury.4. Through female genital tracta. Spontaneousb. Iatrogenic# Iatrogenic perforation of uterus or vagina# Culdocentesis# Tubal patency test5. Gas forming peritonitis6. Pneumothorax with pleuroperitoneal fistula

So the answer to this question is D


Instruction:
Pontic design not indicated in the anterior region is:
A. Spheroidal pontic
B. Ovate pontic.
C. Modified Ridge lap pontic.
D. None.

Response:
None

So the answer to this question is A


Instruction:
Transtuzumab is targeted against?
A. Her 2 neu
B. CD 20
C. EGFR
D. VEGF

Response:
Ans. is 'a' i.e., Her 2 neu

So the answer to this question is A


Instruction:
______________ is not a property of X-rays
A. Ionization
B. Action on photographic film
C. Excitation
D. Collimation

Response:
None

So the answer to this question is D


Instruction:
At distal end of mandibular arch pear shaped structure which is composed of typical gingival tissue known as?
A. Retromolar pad
B. Retromolar papilla
C. Pterygomandibular raphe
D. Masseteric notch

Response:
Retromoplar papilla:

It is an scar/remnants of gingival tissue of last mandibular molar extraction

Retromolar Pad:

Triangular pad of soft tissue at distal end of lower ridge.

So the answer to this question is B


Instruction:
Which is not a complication of chicken pox -
A. Pancreatitis
B. Pneumonia
C. Encephalitis
D. Thrombocytopenia

Response:
Complications

The most common infectious complication of varicella is the secondary bacterial superinfection of the skin, which is usually caused by Streptococcus pyogenes or Staphylococcus aureus.
The most common extracutaneous site of involvement in children is CNS.
Varicella pneumonia is the most serious complication following chickenpox in adults.

So the answer to this question is A


Instruction:
All are true about composite skin graft EXCEPT
A. Covers skins and under tissue
B. High risk failure
C. Skin grafting requires to fill the gap
D. Can be useful in fingertip amputation

Response:
(High risk failure): (7094- CSDT 13th edition; 404-Baily & Love 26th)COMPOSITIVE GRAFTS (usually skin and fat or skin and cartilage)* Free graft that must reestablish its blood supply in the recipient area* Often taken from the ear margin and useful for rebuilding missing elements of nose, eyelids andfingertips* Composite graft must be small or at least relatively thin and will require recipient sites with excellent vasularity* These grafts are generally used in the facesAdvantages and Disdvantages of Various Types of Skin Grafts.Type of GraftAdvantagesDisadvantagesThin split thicknessSurvive transplantation most easily. Donor sites heal most rapidlyFewest qualities of normal skin.Maximum contraction. Least resistance to trauma. Sensation poor. Aesthetically poor.Thick split thicknessMore qualities of normal skin.Less contraction. More resistant to trauma. Sensation fair.Aesthetically more acceptableSurvive transplantation less well.Donor site heals slowlyFull-thicknessNearly all qualities of normal skin.Minimal contraction. Very resistant to trauma. Sensation good. Aesthetically good.Survive transplantation least well.Donor site must be closed surgically. Donor sites are limited.

So the answer to this question is B


Instruction:
Largest protozoan is?
A. E. Histolytica
B. Balantidium coli
C. E.coli
D. Plasmodium

Response:
ANSWER: (B) Balantidium coliREF: Jawett's 24th edition Section VI. Parasitology > Chapter 46-Medical Parasitology"Balantidium coli, the cause of balantidiasis or balantidial dysentery, is the largest intestinal protozoan of humans"

So the answer to this question is B


Instruction:
The most commonly practiced operative procedure for a perforated duodenal ulcer is
A. Vagotomy and pyloroplasty
B. vagotomy and antrectomy
C. Vagotomy and perforation closure
D. Graham's omentum patch repair

Response:
Here perforation is identified and an omental patch is placed & closed with interrupted, horizontal sutures. Because of its adhesion propey, it seals perforation, maintain good vascularity, and promote lymphatic drainage Reference: SRB 5th edition, page no: 832

So the answer to this question is D


Instruction:
In a patient with hypertriglyceridemia and low HDL, which of the following drug will be best without risk of myopathy as side effect ?
A. Fibric acid derivatives
B. Nicotinic acid
C. Atorvastatin
D. Clofibrate

Response:
None

So the answer to this question is B


Instruction:
Which of the following drug is a 5HT2c agonist approved for treatment of obesity and causes Serotonin Syndrome as an adverse effect?
A. Lorcaserin
B. Sibutramine
C. Modafinil
D. Rimonabant

Response:
Lorcaserin: 5HT2c agonist Causes weight loss S/E: Serotonin syndrome Sibutramine: Serotonin and nor-adrenaline reuptake inhibitor Discontinued due to increased risk of stroke Modafinil: used in treatment of Narcolepsy Rimonabant: Inverse agonist of cannabinoid receptor - 1 Causes weight loss Prevents craving of alcohol S/E: psychological problems (withdrawn)

So the answer to this question is A


Instruction:
Which of the following is not associated EBV virus ?
A. IMN
B. Kapasi sarcoma
C. Oral hairy leukoplalkia
D. NHL

Response:
Ans. is 'b' i.e., Kaposi sarcoma

So the answer to this question is B


Instruction:
Radiographically a ragged or poorly defined margin is seen in
A. squamous cell carcinoma
B. osteomyelitis
C. secondarily infected cyst
D. all of the above

Response:
None

So the answer to this question is D


Instruction:
The egg of which helminth can be concentrated in saturated salt solution -
A. Taenia saginata
B. Taenia solium
C. Unfeilized egg of Ascaris
D. Ancylostoma duodenale

Response:
Ans. is 'd' i.e., Ancylostoma duodenale

So the answer to this question is D


Instruction:
A one month old baby presents to the emergency room with history of vomiting, irritability and jaundice. The baby was previously healthy without any symptoms. One week ago cereals and fruit juices were introduced into the diet. USG revealed hepatomegaly and liver functions test were abnormal. The deficient enzyme is
A. Fructokinase
B. Aldolase-B
C. Galactose-1 phosphate uridyl transferase
D. Acid alpha glucosidase

Response:
Answer: b) Aldolase-B (NELSON 20th ED, p-727)HEREDITARY FRUCTOSE INTOLERANCEDeficiency of fructose-1,6-bisphosphate aldolase (aldolase B)Patients are asymptomatic until fructose or sucrose (table sugar) is ingested (usually from fruit, fruit juice, or sweetened cereal)Symptoms may occur early in life, soon after birth if foods or formulas containing these sugars are introduced into the diet.Early clinical manifestations resemble galactosemia and include jaundice, hepatomegaly, vomiting, lethargy, irritability, and convulsionsLaboratory findings include prolonged clotting time, hypoalbuminuria, elevation of bilirubin and transaminase levels, and proximal tubular dysfunctionTreatment - complete elimination of all sources of sucrose, fructose, and sorbitol from the dietGalactose-1-phosphate uridyl transferase deficiency - Onset of symptoms typically by the second half of the 1st wk of lifeFructokinase deficiency - not associated with any clinical manifestations.Lysosomal acid alpha glucosidase (Pompe's disease) - affected infants present in the first few weeks to months of life with hypotonia, generalized muscle weakness with a "floppy infant" appearance, neuropathic bulbar weakness, feeding difficulties, macroglossia, hepatomegaly, and a hypertrophic cardiomyopathy followed by death from cardiorespiratory failure or respiratory infection usually by 1 yr of age

So the answer to this question is B


Instruction:
A 27 year old man with AIDS develops a reddish, slightly raised rash on his face, neck, and mouth, consistent in appearance with Kaposi's sarcoma. Microscopically, the proliferating cells in this malignancy most closely resemble which of the following?
A. Angiosarcoma
B. Carcinosarcoma
C. Lymphoma
D. Malignant fibrous histiocytoma

Response:
Kaposi's sarcoma is a spindle cell neoplasm that is highly associated with AIDS and with the Herpes simplex virus type 8. The tumor has an appearance very similar to that of angiosarcoma-proliferating stromal cells and endothelium creating vascular channels that contain blood cells. Carcinosarcoma is a tumor that contains malignant epithelial cells and malignant stromal cells. There is no epithelial element in Kaposi's sarcoma. Although lymphoma occurs with increased frequency in AIDS, it does not resemble Kaposi's sarcoma. Lymphoma involves neoplastic lymphocytes, whereas Kaposi's sarcoma involves neoplastic vascular structures. Malignant fibrous histiocytoma (MFH) is an extremely poorly differentiated (anaplastic) stromal malignancy. MFH does not produce any recognizable mesenchymal structures- thus, the production of vascular structures by Kaposi's sarcoma differentiates the two tumors. Ref: Wyatt C., Butterwoh IV J.F., Moos P.J., Mackey D.C., Brown T.G. (2008). Chapter 9. Vascular Pathology. In C. Wyatt, J.F. Butterwoh IV, P.J. Moos, D.C. Mackey, T.G. Brown (Eds), Pathology: The Big Picture.

So the answer to this question is A


Instruction:
Preterm infant is defined as -
A. Born between 37-42 weeks
B. Born > 42 weeks
C. Born < 37 weeks
D. Born <25 weeks

Response:
Preterm is defined as babies born alive before 37 weeks of pregnancy are completed. Subcategories include- 1. Extremely preterm is less than 28 wks 2. Very preterm 28 to 32 wks 3. Moderate to late preterm 32 to 37 wks Reference Park&;s textbook of preventive and social medicine 23rd edition pg 536

So the answer to this question is C


Instruction:
The following condition is not associated with an Anti phospholipid syndrome -
A. Venous thrombosis
B. Recurrent foetal loss
C. Thrombocytosis
D. Neurological manifestation

Response:
The clinical manifestations of APLA syndrome includes adverse pregnancy outcome (recurrent first trimester aboion ),venous thromboembolism,aerial thromboembolism,livedo reticularis,transverse myelitis,skin necrosis,chorea.Davidson's principles &practice of medicine 22nd edition pg no 1055.

So the answer to this question is C


Instruction:
The drug which causes renal papillary necrosis is?
A. Acetophenetidin
B. Gentamicin
C. Gold
D. Methicillin

Response:
Ans. (a) AcetophenetidinRef: Robbiris pathology 9th ed. /423. 936t* Renal papillary necrosis due to acetophenetidin/ phenacetin is a gradual process with slow loss of renal function.* The most common cause of renal papillary necrosis is diabetic glomerulopathy

So the answer to this question is A


Instruction:
Only definitive indication of systemic steroids in psoriasis is
A. Pustular psoriasis
B. Erythroderma
C. Psoriatic ahropathy
D. Impetigo herpetiformis

Response:
D i.e. Impetigo herpetiformis - Systemic steroids should be avoided in routine care of psoriasis Q because the disease usually breaks through, requiring progressively higher doses to control symptoms and withdrawl of drug is usually associated with frequent relapse in form of life threatening erythrodermic psoriasis (with exfoliative dermatitis) and pustular psoriasis (with pus lakes)Q. - However, systemic steroids may have a role in management of Persistent, otherwise uncontrollable (e.g. with metabolic complications), psoriatic erythroderma and in fulminant generalized pustular psoriasis (von Zumbusch type) if other drugs are ineffective or contraindicatedQ. Pustular psoriasis in pregnancy is called impetigo herpatiformis. In pregnancy, safest drug for treatment of pustular psoriasis is prednisoloneQ. Severe psoriatic polyahritis threatening severe irreversible joint damageQ. Pustular psoriasis may develop after strong topical or systemic steroids have been used and then abruptly withdrawnQ. It presents with development of generalized pustules all over the bodyQ. Erythematous skin rash with multiple pus lakes suggests a diagnosis of generalized pustular psoriasis. As retinoids, methotrexate & PUVA cannot be used (or contraindicated) in pregnancy. Coicosteroids form the mainstay of treatment for generalized pustular psoriasis in pregnancyQ. Localized disease is best treated with topical steroid & fulminating generalized disease is best treated with systemic prednisolone (oral).

So the answer to this question is D


Instruction:
Kit 3 is given for :
A. Urethral discharge
B. Genital ulcer (herpetic)
C. Genital ulcer (non herpetic)
D. Genital ulcer (non herpetic), allergic to penicillin

Response:
Kit 3(white) is given for Non-herpetic genital ulcer.

So the answer to this question is C


Instruction:
Bradycardia is seen with:
A. Midazolam
B. Epinephrine
C. Succinyl choline
D. Dopamine

Response:
Ans: C (Succinyl choline) Ref: Ajay Yadav, Short textbook of Pediatrics, 1st editionExplanation:Cardiac Sideeffects of Anesthetic DrugsDrugsCardiac Side effectsMidazolamMinimal reduction in Heart rate, Blood pressure and cardiac outputEpinephrineProduces tachycardia, hypertension and ventricular arrhythmiasSuccinyiCholineIt produces muscarinic effects, similar to Acetyl cholineIt causes Profound BRADYCARDIA, so atropine should be given prior to use of Succinylcholine Choline, especially in childrenDopamineProduces tachycardia, hypertension and ventricular arrhythmiasThiopentoneit causes Hypotension, which is more because of venodilatation and direct depression of vasomotor centreDirect myocardial depressant

So the answer to this question is C


Instruction:
Epstein-Barr virus causes all the following EXCEPT:
A. Burkitt's lymphoma
B. Nasopharyngeal carcinoma
C. Carcinoma cervix
D. Infectious mononucleosis

Response:
Ans. c (Carcinoma cervix). (Ref. Harrison, 17th/Table 105-4; Microbiology by Anantanaray an, 6th/448). Infectious AgentLymphoid Malignancy1.Epstein-Barr virus (EBV)Burkitt's lymphomaPost-organ transplant lymphomaPrimary CNS diffuse large B cell lymphomaHodgkin's diseaseExtranodal NK/T cell lymphoma, nasal type2.HTLV-IAdult T cell leukemia/lymphoma3.HIVDiffuse large B cell lymphomaBurkitt's lymphoma4.Hepatitis C virusLymphoplasmacytic lymphoma5.Helicobacter pyloriGastric MALT lymphoma6.Human herpesvirus 8Primary effusion lymphomaMulticentric Castleman's diseaseEducational point:# Carcinoma cervix is supposed to be caused by HPV 16, 18, 31, 33.

So the answer to this question is C


Instruction:
True about the chronic obstructive pulmonary disease (COPD)
A. FEVI <30 of predicted value
B. FEV1/FVC<0.7
C. total lung capacity increased
D. All

Response:
In COPD, airflow obstruction is demonstrated by spirometry in which the post-bronchodilator FEV1/FVC is less than 70%.
The severity of COPD may be defined in relation to the post-bronchodilator FEV1 which is <30% predicted in very severe type.
Measurement of lung volumes provides an assessment of hyperinflation.COPD is characterised by an increase in residual volume, residual volume/total lung capacity ratio, and progressive hyperinflation(increased total lung capacity) late in the disease.

So the answer to this question is D


Instruction:
All are involved in non toxic transpo of ammonia except
A. Glutamine synthetase
B. Glutaminase
C. Alanine cycle
D. SGPT

Response:
The amino acid used primarily by muscle, for transpo of ammonia to liver is alanine. Most other tissues however use glutamine

So the answer to this question is B


Instruction:
Gall stone causes intestinal obstruction when it gets impacted in which part of the intestine commonly
A. Proximal ileum
B. Distal ileum
C. Duodenum
D. Jejunum

Response:
This type of obstruction tends to occur in the elderly secondary to erosion of a large gallstone through the gall bladder into the duodenum. Classically, there is impaction about 60 cm proximal to the ileocaecal valve. The patient may have recurrent attacks as the obstruction is frequently incomplete or relapsing as a result of a ball-valve effect.

So the answer to this question is B


Instruction:
HCN causes -
A. Histotoxic anoxia by inhibiting carbonic anhydrase
B. Histotoxic anoxia by inhibiting succinyl oxidase
C. Cytotoxic anoxia by inhibiting cytochrome oxidase
D. Cytotoxide anoxia by inhibiting succinyl oxidase

Response:
Hydrocyanic acid Cyanide occurs as solid, liquid or in gaseous state as: * Gas form -- Hydrogen cyanide (HCN) * Liquid form -- Hydrocyanic acid (Prussic acid) * Solid form -- occurs as salts such as potassium cyanide/ sodium cyanide. Mechanism of Action Cyanide reversibly inhibits ferric iron containing enzymes. Cyanide attaches to the iron of the prosthetic group of cytochrome oxidase resulting in disturbance of the transpo and utilization of oxygen in cells and causing a cytotoxic anoxia. Fatal Dose * 50 to 100 mg for hydrocyanic acid * 200 to 300 mg for sodium/potassium cyanide * 50 to 80 bitter almonds * Inhalation of 1 pa in 2000 -- hydrogen cyanide Fatal period: 2 to 10 minutes. Clinical Features Inhalation * Constriction about throat * Dizziness * Loss of consciousness * Coma * Death Ingestion * CNS: Headache, anxiety, agitation, dizziness, confusion, convulsions, coma * CVS: Initially bradycardia and hypeension and latter tachycardia and hypotension, arrhythmias * RS: Tachypnea followed by bradypnea * GIT: Nausea, vomiting, abdominal pain, numbness. Skin * Perspiration * Bullae. Chronic poisoning: Chronic low level of exposure leads to: * Headache * Amblyopia * Optic atrophy * Peripheral neuropathy * Ataxia * Deafness * Glossitis * Stomatitis. Management * Ingestion: gastric lavage with 5 percent sodium thiosulfate solution * Antidote- amyl nitrate -- inhaled * Sodium nitrate slowly I.V. * Sodium thiosulfate 25 percent solution I.V. * Mechanism of action of nitrites: nitrates induce methemoglobinemia, which causes detachment of cyanide from the heme group of cytochrome oxidase since methemoglobin has a higher binding affinity for cyanide. Cyanide combines with methemoglobin and form non-toxic cynmethemoglobin. * Mechanism of action of sodium thiosulfate: sodium thiosulfate serves as a substrate for the enzyme rhodanese to catabolise cyanide to non-toxic thiocyanate, which is excreted in the urine. Autopsy Findings * Bitter almond like smell * Cherry red color of postmoem lividity * Cyanosis * Froth at mouth and nose * Bright red color blood9 * Pulmonary edema * Serosal surface may show hemorrhages * Stomach mucosa may be eroded and blackened due to formation of alkaline hematin. Samples to be preserved for chemical analysis * Blood * Routine viscera * Lung * Brain * Hea * Spleen is considered as best specimen due to presence of more RBCs. REF;THE SYNOPSIS OF FORENSIC MEDICINE:KS NARAYANA REDDY;28th EDITION;PAGE NO 362

So the answer to this question is C


Instruction:
Which of the following is true of fat embolism -
A. Thrombocytopenia
B. Macroglobulinemia
C. Prothrombin time increased
D. Fat globules in urine

Response:
symptomatic fat embolism syndrome characterized by pulmonary insufficiency, neurologic symptoms, anemia, thrombocytopenia, and a diffuse petechial rash, which is fatal in 10% of cases. Clinical signs and symptoms appear 1 to 3 days after injury as the sudden onset of tachypnea, dyspnea, tachycardia, irritability, and restlessness, which can progress rapidly to delirium or coma ( Robbins Basic Pathology, 9 th edition, page 91 )

So the answer to this question is A


Instruction:
The ratio of the dose that produces toxicity to the dose that produces a clinically desired or effective response in a population of individual is known as:DELETE REPEAT
A. Efficacy
B. Potency
C. Therapeutic index
D. Paial agonist

Response:
ref ; KD Tripathi pharmacology 7th edition (page no; 55) The gap between the therapeutic effect DRC (Dose Response Curve) and the adverse effect DRC defines the safety margin or the therapeutic index of a drug. Therapeutic index =median lethal dose /median effective dose or LD 50/ ED 50

So the answer to this question is C


Instruction:
Resonance of tympanic membrance is
A. 800 Hz
B. 800-1600 Hz
C. 3000 Hz
D. None

Response:
External auditory canal 3000 Hz Tympanic membrane 800-1600 Hz Middle ear 800 Hz Ossicular chain 500-2000 Hz Ref: Dhingra 7e pg 17.

So the answer to this question is B


Instruction:
The external urethral sphincter is located in which of the following regions?
A. Ischiorectal fossa
B. Extrapeoneal space
C. Retropubic space
D. Deep perineal space

Response:
The external urethral sphincter is the voluntary sphincter composed of skeletal muscle, the sphincter urethrae muscle. This muscle, along with the deep transverse perineal muscle, are in the deep perineal space and form the muscular layer of the urogenital diaphragm. The internal urethral sphincter is composed of smooth muscle in the neck of the bladder. The internal and external sphincters are both open during micturition. During ejaculation, the external sphincter is open and the internal sphincter is closed.

So the answer to this question is D


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

Response:
Thyroid cancers are classified according to their histopathological characteristics and they are: ? Papillary thyroid cancer (75% to 85% of cases ) - often in young females - excellent prognosis ? Follicular thyroid cancer (10% to 20% of cases ) ? Medullary thyroid cancer (5% to 8% of cases)- cancer of the parafollicular cells, often pa of MEN-2 syndrome. ? Anaplastic thyroid cancer (Less than 5%). It is not responsive to treatment and can cause pressure symptoms. ? O thers ? ? Lymphoma ? ? Squamous cell carcinoma, sarcoma Ref: Robbin's Basic Pathology, 7th Edition, Page 1177; Harrison's Internal Medicine, 18th Edition, Page 2243; Bailey and Love's Sho Practice of Surgery, 25th Edition, Page 792

So the answer to this question is C


Instruction:
A 35 year old alcoholic presents with diarrhoea and scaly rash on his neck, hands and feet. He will improve if he is treated with which one of the following ?
A. VitaminA
B. Thiamine
C. Folic acid
D. Niacin

Response:
*Niacin encompasses nicotinic acid and nicotinamide . *Niacin deficiency causes pellagra - diarrhea,dermatitis, dementia,death Niacin encompasses nicotinic acid and nicotinamide. Nicotinamide is an essential pa of the two pyridine nucleotides, nicotinamide adenine dinucleotide (NAD) and nicotinamide adenine dinucleotide phosphate (NADP), which play a key role as hydrogen acceptors and donors for many enzymes. Niacin can be synthesised in the body in limited amounts from the amino acid tryptophan. Deficiency - pellagra Pellagra was formerly endemic among poor people who subsisted chiefly on maize, which contains niacytin, a form of niacin that the body is unable to utilise. Pellagra can develop in only 8 weeks in individuals eating diets that are very deficient in niacin and tryptophan. It remains a problem in pas of Africa, and is occasionally seen in alcoholics and in patients with chronic small intestinal disease in developed countries. Pellagra can occur in Hanup's disease, a genetic disorder characterised by impaired absorption of several amino acids, including tryptophan. It is also seen occasionally in carcinoid syndrome (p. 678), when tryptophan is consumed in the excessive production of 5-hydroxytryptamine (5-HT, serotonin). Pellagra has been called the disease of the three Ds: * Dermatitis. Characteristically, there is erythema resembling severe sunburn, appearing symmetrically over the pas of the body exposed to sunlight, paicularly the limbs and especially on the neck but not the face (Casal's necklace, Fig. 19.13). The skin lesions may progress to vesiculation, cracking, exudation and secondary infection. * Diarrhoea. This is often associated with anorexia, nausea, glossitis and dysphagia, reflecting the presence of a non-infective inflammation that extends throughout the gastrointestinal tract. * Dementia. In severe deficiency, delirium occurs acutely and dementia develops in chronic cases. Treatment is with nicotinamide, given in a dose of 100 mg 3 times daily orally or parenterally. The response is usually rapid. Within 24 hours the erythema diminishes, the diarrhoea ceases and a striking improvement occurs in the patient's mental state. Ref Harrison 20th edition pg 98e

So the answer to this question is D


Instruction:
Commonest site of feilization is :
A. lsthmic
B. Ampulla
C. lnfundibulum
D. lnterstitial

Response:
Ampulla

So the answer to this question is B


Instruction:
500 people of viral fever are treated with a medicine. 80 percent were cured in 3 days. True statement about efficacy of medicine -
A. Medicine is effective
B. Cannot be commented
C. Medicine is ineffective
D. None of the above

Response:
None

So the answer to this question is B


Instruction:
Which Rectus muscle is not supplied by occulomotor nerve ?
A. Medial rectus
B. Lateral rectus
C. Superior rectus
D. Inferior rectus

Response:
Ans. is 'a' i.e., Right side TestisTestis is the male gonad. It is suspended in the scrotum by spermatic cord.Left testis is slightly (1 cm) lower than the right one.The average testicular dimensions are 5 cm (length), 25 cm (breath), 3cm (anteroposterior diameter or thickness) and 10.5-14 gms (weight).The epididymis is attached to is posterolateral surface.Vas (ductus) deference arises from lower pole of epididymis.Lateral pa of epididymis is separated from testis by an extension of cavity of tunica vaginalis, called sinus of epididymis.There is small oval body attached to upper pole of testis called appendix of testis, a remnant of paramesonephric duct.Testis is covered by three coats (from outside to inside) :- (i) tunica vaginalis (which has a parietal and a visceral layer); (ii) tunica albuginea; and (iii) tunica vasculosa. Outside these three coats, there are layers of scrotum. StructureTestis consists of 200-300 lobules.Each lobules consists of 2-3 seminiferous tubules (where spermatozoa are formed).Seminiferous tubules join at the apex of lobules to form straight tubules, which anastomose with each other to form rete testis.Which emerge from upper pole and enter the epididymis. Each tubule becomes highly coiled to form a lobe of the head of epididymis.All tubules end in a single duct which is coiled on itself to form the body and tail of the epididymis. It is continuous with ductus deference.Aerial supplyTestes are supplied by testticular aery, a branch of abdominal aoa, arises at L2 veebral level. Venous drainageThe veins emerging from the testis and epididymis form the pampiniform plexus.Anterior pa of the plexus is arranged around the testicular aery, middle pa around ductus deferens and its aery, and posterior pa is isolated.The plexus condenses into 4 veins at the superficial inguinal ring and 2 veins at the deep inguinal ring. Ultimately, one vein is formed which drains into inferior vena cava on the right and into left renal vein on the left side.Lymphatic drainageLymphatics from the testis run back with the testicular aery to para-aoic nodes along side the aoa at the level of origin of testicular aeries (L2 veebra).

So the answer to this question is A


Instruction:
Most common type of twin pregnancy is-
A. veex + transverse
B. Both veex
C. Veex + breech
D. Both breech

Response:
<p> Most common type of presentation in twin pregnancy is both twin veex(50%) Reference:DC Dutta&;s textbook of obstetrics,8th edition, page no:243 <\p>

So the answer to this question is B


Instruction:
A child comes with steroid resistant nephrotic syndrome secondary to FSGS, not responsive to methylprednisolone. What next should be given
A. Oral cyclophosphamide
B. Oral cyclosporine
C. Oral mycophenolate
D. IV cyclophosmaphid

Response:
When child with nephrotic syndrome is restaurant to steroids then calcineurin inhibitors such as cyclosporine and tacrolimus the first drug of choice. Reference: GHAI Essential pediatrics, 8th edition

So the answer to this question is B


Instruction:
Decreased ability to smell is
A. Anosmia
B. Hyposmia
C. Phantosmia
D. Hyperosmia

Response:
(B) Hyposmia # Hyposmia: Hyposmia is a state of decreased sensation of smell> Disorders of the Sense of Smell: Transport losses - Rhinitis, polyps/neoplasms of nasal cavity Sensory losses - Drugs, radiation therapy, toxin exposure Neural losses - Alps, alcoholism, Alzheimer's, Kallman syndrome, hypothyroidism etc.,> Treatment: There is no proven treatment for sensorineural olfactory losses. Fortunately, spontaneous recovery often occurs.> Zinc and vitamin therapy (especially with vitamin A) are advocated by some.> Profound zinc deficiency can produce loss and distortion of the sense of smell but is not a clinically important problem except in very limited geographic areas

So the answer to this question is B


Instruction:
Normal color of healthy gingiva in children:
A. Pink
B. Red
C. Coral pink
D. Pale (light) pink

Response:
None

So the answer to this question is D


Instruction:
Ifosfamide belongs to which Group of anticancer drugs
A. Alkylating agent
B. Antimebolites
C. Mitotic inhibitors
D. Topoisomerase inhibitors

Response:
Refer principles of Pharmacology 1st /869 KDT 6/e p 819 Ifosfamide is an alkyatingg agent

So the answer to this question is A


Instruction:
Whipple triad is seen in:
A. Glucogonoma
B. Insulinoma
C. Gastrinoma
D. Ca pancreas

Response:
Ans: b (Insulinoma) Ref: Bailey, 24th ed, p. 1131Whipple triad is the classic diagnostic criteria for insulinoma.Whipple triad- Signs & symptoms of hypoglycaemia during fasting.- Blood glucose below 2.8 mmol/ltr (50 mg/dl) during symptomatic episodes.- Relief of symptoms by administration of glucose.Insulinomass-cell endocrinal tumours of pancreasArise from whole pancreasSymptoms are due to excessive secretion of insulin causing hypoglycaemia.Most are solitary & benign.Only 10- 15% are malignant.LocalizationMore sensitive test is intraoperative USGHigh resolution CT & MRIEndoscopic USGSelective arteriographyTreatmentSurgical excision is treatment of choice.

So the answer to this question is B


Instruction:
All except one is true regarding hyoid bone
A. Hyoid bone lies at level of 3rd cervical vertebra
B. It develops from 3rd  and 4rd  branchial arches
C. Fracture of hyoid bone strongly suggest strangulation
D. Lesser cornea provides attachment to stylohyoid ligament

Response:
Hyoid bone develops from 2nd and 3rd branchial arches.

So the answer to this question is B


Instruction:
Advantage of using chemically activated resin for repairing is:
A. They can be polymerized at room temperature
B. They are easy to manipulate
C. Easy to handle
D. Cheap in cost

Response:
ADVANTAGES AND DISADVANTAGES

Better initial fit, which is because the curing is carried out at room temperature. Thus there is less thermal contraction.
Color stability is inferior to that of heat cure resin, due to subsequent oxidation of the tertiary amine.
Slightly inferior properties because the degree of polymerization of self curing acrylics is less than that of heat cured ones.
For repairing dentures, self curing resins are preferable to heat cured resins as heat curing causes warpage.

So the answer to this question is A


Instruction:
All the following affects absorption of calcium, EXCEPT
A. Citric acid
B. Retinoic acid
C. Phytates
D. Oxalates

Response:
None

So the answer to this question is B


Instruction:
Low dose radiotherapy is given for:
A. Seminoma
B. Malignant melanoma
C. Osteosarcoma
D. Chondrosarcoma

Response:
Ans. Seminoma

So the answer to this question is A


Instruction:
Lymphohistocytic variant of Reed Sternberg seen in which subtype of Hodgkin lymphoma:
A. Nodular sclerosis
B. Lymphocyte rich
C. Lymphocyte predominant
D. Lymphocyte depleted

Response:
Ref. Robbins Pathology. 9th edition. Page. 601
Direct picture from Robbins

So the answer to this question is C


Instruction:
Major End Product of fatty acid synthesis is:
A. Acetyl CoA
B. ATP
C. Citrate
D. Palmitate

Response:
Ref. Textbook of Biochemistry for Medical Students. Page. 198

Excess dietary glucose can be converted to fatty acids in the liver and subsequently sent to the adipose tissue for storage.
Adipose tissue synthesizes smaller quantities of fatty acids.
Insulin promotes many steps in the conversion of glucose to acetyl CoA in the liver:

Glucokinase (induced)

PFK-2/PFK-1 (PFK-2 dephosphorylated)

Pyruvate dehydrogenase (dephosphorylated)
Both of the major enzymes of fatty acid synthesis are also affected by insulin:

Acetyl CoA carboxylase (dephosphorylated, activated)
Fatty acid synthase (induced)

Final end product is Palmitate, after which fatty acid elongation happens. Look at the belwo diagram for steps

So the answer to this question is D


Instruction:
Deficiency of Copper causes
A. Anemia
B. Wilson's disease
C. Hypothyroidism
D. Obtundation

Response:
(A) Anemia

So the answer to this question is A


Instruction:
Child with "Intracranial Calcification, Chorioretinitis" most appropriate cause is:
A. Toxoplasmosis
B. Herpes simplex
C. CMV
D. Syphilis

Response:
Toxoplasmosis

So the answer to this question is A


Instruction:
Mutually protected occlusion; TRUE is
A. Is an occlusal scheme in which the posterior teeth prevent excessive contact of the anterior teeth in maximum intercuspation
B. Posterior teeth function most effectively in stopping the mandible during closure
C. Anterior teeth function most effectively in guiding the mandible during eccentric movements
D. All of the above

Response:
None

So the answer to this question is D


Instruction:
Which of the following product is primarily filtered with little secretion or re - absorption in renal tubules?
A. Sodium
B. Creatinine
C. Glucose
D. Amino acids

Response:
Ans. (b) Creatinine(Ref: Ganong, 25th ed/p.679)Substances like inulin and creatinine are useful to estimate GFR because they are freely filtered, neither resorbed, nor secreted

So the answer to this question is B


Instruction:
The leukotriene receptor antagonist used in bronchial asthma is:
A. Zafirleukast
B. Zileuton
C. Ketotifen
D. Omalizumab

Response:
Among the options given only Zafirleukast is leukotriene receptor antagonist. Zileuton is a 5-Lipoxygenase inhibitor, Ketotifen is an antihistaminic (H1) and Omalizumab is a recombinant humanized monoclonal anti-IgE antibody. Ref: Bronchial Asthma: A Guide for Practical Understanding and Treatment, Springer, 2011, Page 193; Current Review of Asthma, By Michael A. Kaliner - Page 216; Essentials of Medical Pharmacology, KD Tripathi - 5th Edition, Page 205-206; Katzung & Trevor's Pharmacology 10th Edition, Page 325.

So the answer to this question is A


Instruction:
Antral puncture is done through:
A. Superior meatus
B. Inferior meatus
C. Middle meatus
D. All

Response:
In proof puncture or antral lavage, the lateral wall of inferior meatus is punctured with Lichtwitz trocar and cannula. This site is selected as bone is very thin here.

So the answer to this question is B


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

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

So the answer to this question is A


Instruction:
If 25 persons are working on a project for 30 years. There is how many person-years of employement.
A. 75
B. 750
C. 120
D. 1200

Response:
Ans. is 'b' i.e.,750 "10000 people working for 1 year would lead to 1000 person years of employment"Thus, 25 persons working for 30 years will lead to 750 person-years of employment.

So the answer to this question is B


Instruction:
Which of the following metabolic complications is not seen in Chronic Renal Failure:
A. Hyperkalemia
B. Hyponatremia
C. Hypercalcemia
D. Hyperphosphatemia

Response:
Answer is C (Hypercalcemia) Chronic Renal Failure is associated with Hypocalcemia and not hypercalcemia.

So the answer to this question is C


Instruction:
Which of the following is primary cell line ?
A. Chick embryo fibroblast
B. Hela cells
C. Vero cells
D. WI-38

Response:
Ans. is 'a' i.e., Chick embryo fibroblast Chick embryo fibroblast is primary cell culture.

So the answer to this question is A


Instruction:
Drug Induced Lupus is caused by all except
A. Sulphonamides
B. Hydralazine
C. Isoniazid
D. Procaine

Response:
Drugs metablized by acetylation can cause drug induced lupus: "D.SHIP"

Dapsone
Sulphonamides
Isoniazid
Procainamide (not Procaine).

So the answer to this question is D


Instruction:
The voice in a patient with Bilateral abductor paralysis of the larynx is
A. Puberphonia
B. Phonasthenia
C. Dysphonia plicae ventricularis
D. Normal or good voice

Response:
The vocal cord vibrates during speech to produce speech. These vibrations occur due to subglottic airflow when the cord is in adduction (Median & paramedian).
During normal phonation the vocal cords are in median position and during strong whisper vocal cords are in paramedian position.
So, for proper voice, the vocal cords should be in median or paramedian position.
Same occurs in bilateral recurrent laryngeal nerve paralysis, where the vocal cords are in median or paramedian position → Voice is less likely to be impaired.

Now, the question comes to the mind immediately : - Why the voice is affected in unilateral paralysis, in which the cord is in median or paramedian position. The answer is : -

In unilateral vocal cord paralysis, one cord is paralysed and vibrates abnormally or does not vibrate at all, therefore cannot coordinate with the other side (normal) vocal cord. This results in voice problems.
In bilateral paralysis, both vocal cords vibrate abnormally but coordinate with each other, therefore voice is less affected.

So the answer to this question is D


Instruction:
All of the following are true for retinopathy of prematurity except:
A. Occurs in premature infants due to late crying
B. Due to hypoxia there occurs neovascularization followed by fibroproliferation
C. End result is bilateral blindness
D. Blindness can be prevented by early diagnosis and ablation of vascular premature retina with cryotherapy or photocoagulation

Response:
Ans. Occurs in premature infants due to late crying

So the answer to this question is A


Instruction:
Primary ciliary dyskinesia is associated with all except -
A. Sterility in males
B. Hypothyroidism
C. Sinusitis
D. Respiratory infection

Response:
Ans. is 'b' i.e., Hypothyroidism Kartagener's syndrome is characterized bv triad ofo Situs inversuso Chronic sinusitiso BronchiestasisIt is a subset of primary ciJiarv dyskinesiaPrimary ciliary dyskinesiao Primary ciliary dyskinesia (PCD, also called the immotile-cilia syndrome) is characterized by congenital impairment of mucociliary clearance (MCC).o Clinical manifestations includeChronic coughChronic rhinitis andChronic sinusitisOtitis and otosalpingitis (common in childhood)Nasal polyposis and agenesis of ike frontal sinusesClinical features of primary ciliary dyskinesiaPulmonaryRhinosinusitisOtitisCentral nervous systemFertilityo Mild respiratory distresso Runny nose ando Chronic secretoryo Chronic sinusitiso Infertilityo Increased incidence ofround the yearotitis media withcauses fatigue and(immoiilerespiratory infectionsnasal congestionrecurrent episodesheadachespermatozoawith chronic cough ando Nasal polyposisof acute otitis mediao Hydrocephalus duemotileexpectoration of Sitns inversusto impairedspermatozoa),mucopurluent sputum o Complete reversalfunction of o Bronchiestasis of the circulatoryependymal cilia | system and viscera Associated abnormalities o No adverse health o A number of congenital consequence abnormalities are : * Transposition of great vessels * Pyloric stenosis * Epispadias

So the answer to this question is B


Instruction:
Succinyl Co-A is formed by:
A. Histidine
B. Leucine
C. Valine
D. Lysine

Response:
Succinyl COA is formed by Valine, Isoleucine, Methionine.

So the answer to this question is C


Instruction:
What is false about pertussis -a) Maternal antibody provides protection in infantsb) Fomites play a small role in spread of diseasec) Commonly seen in infantsd) Males and females both are equally affected
A. b
B. ad
C. ac
D. ab

Response:
None

So the answer to this question is B


Instruction:
All are causes of prolapse of cervix EXCEPT: March 2013
A. Many MTP
B. Chronic cough
C. Menopause
D. Delivery of a big baby

Response:
Ans. A i.e. Many MTP

So the answer to this question is A


Instruction:
Maximum incidence of impotance is seen in the patients with following anti hypeensive drug
A. CCBs
B. Beta blockers
C. ARBs
D. ACE inhibitor

Response:
Common side effects of beta blockers include: Dizziness. Weakness. Drowsiness or fatigue. Cold hands and feet. Dry mouth, skin, or eyes. Headache. Upset stomach. Diarrhea or constipation. Refer kDT 6/e p139

So the answer to this question is B


Instruction:
Coronary Ligament is present in:
A. Shoulder Joint
B. Hip Joint
C. Knee Joint
D. Ankle Joint

Response:
Ans. C. Knee JointThe coronary ligaments of the knee (also known as meniscotibial ligaments) are portions of the joint capsule which connect the inferior edges of the fibrocartilaginous menisci to the periphery of the tibial plateaus.

So the answer to this question is C


Instruction:
The antidote of poisoning due to Arninanita muscaria is
A. Physostigmine
B. Amyl nitrate
C. Methylene blue
D. Atropine

Response:
D i.e. Atropine Amanita muscaria & Amanita phalloides are poisonous mushrooms with parasympathetic (Cholinergic) symptoms. So AtropineQ is used as antidote in severe cases; but avoided in mild case as it excerhates delirium.

So the answer to this question is D


Instruction:
Which of the following structure in spermatic cord doesn't come during surgery for vasectomy?
A. Testicular aery
B. Ilioinguinal nerve
C. Autonomic nerves
D. Testicular vein

Response:
The ilioinguinal nerve (L1) passes through pa of the inguinal canal but not a pa of spermatic cord.A vasectomy is a simple procedure in that it only requires an incision through the skin and layers of each spermatic cord to cut the ductus deferens. The spermatic cord contains the same muscular and fascial layers as the anterior abdominal wall because the testis descends through the abdominal wall to the scrotum during embryonic development. The spermatic cord contains the following structures:Ductus deferensTesticular (gonadal) aeryTesticular (gonadal) veinAutonomic neuronsRef: Moon D.A., Foreman K.B., Albeine K.H. (2011). Chapter 7. Anterior Abdominal Wall. In D.A. Moon, K.B. Foreman, K.H. Albeine (Eds), The Big Picture: Gross Anatomy.

So the answer to this question is B


Instruction:
A patient sustained a hyperextension injury to his knee. Which of the following ligament prevents excess - anterior gliding movement of femur on tibia?
A. Anterior cruciate ligament
B. Posterior cruciate ligament
C. Medial collateral ligament
D. Lateral collateral ligament

Response:
*ACL- prevents anterior gliding of tibia on femur *PCL- prevents posterior gliding of tibia on femur Ref: spos medicine 4th/e p.83

So the answer to this question is B


Instruction:
Wolffian duct structures do not develop in all except
A. Swyers syndrome
B. Testicular feminization syndrome
C. Incomplete AIS
D. Turners syndrome

Response:
In incomplete Androgen insensitivity syndrome, there is little responsiveness to testosterone.
So Wolffian duct derivatives are present.

So the answer to this question is C


Instruction:
All of the following are components of Horner's syndrome except -
A. Ptosis
B. Exophthalmos
C. Anhidrosis
D. Loss of cilio- spinal reflex

Response:
None

So the answer to this question is B


Instruction:
Which of the following is not a reducing sugar?
A. Fructose
B. Galactose
C. Sucrose
D. Maltose

Response:
Sucrose contains glucose and fructose .sucrose is not a reducing sugar, and it will not form osazone. is because the linkage involves first carbon of glucose and second carbon of fructose, and free reducing groups are not available.Ref: DM Vasudevan, 7th edition, page no: 77

So the answer to this question is C


Instruction:
All of the following vitamins are involved in sulphur containing amino acid metabolism except?
A. Vitamin B3
B. Vitamin B6
C. Vitamin B9
D. Vitamin B12

Response:
Vitamin B6, B9 and B12 are involved in sulphur containing amino acid metabolism. Vitamin B6 (Pyridoxine) is the co-enzyme of cystathionine b synthase. Vitamin B9 (Folic acid) and B12 (Cobalamin) are involved in the conversion of homocysteine to methionine by methionine synthase. Clinical impoance: Reference: Harpers illustrated biochemistry 30th edition

So the answer to this question is A


Instruction:
Good prognostic factor for AML are all except -
A. Age less than 2 years
B. Acute promyelocytic leukemia
C. Acute megakaryoblastic leukemia
D. Associated with down syndrome

Response:
Ans. is 'c' i.e., Acute Mega karyoblastic leukemia Good Prognostic factors for children with AMLo Prognostic factors are not quite as important in predicting outcome or in guiding treatment for AML as they* are for ALL.o Age at diagnosis: Children younger than age 2.o Initial white blood cell (WBC) count; Children with AML whose WBC count is less than 100,000 cells per cubic millimeter Down syndrome: Children with Down syndrome who develop AML tend to have a good outlook.o Subtype of AML: the acute promyelocytic leukemia (APL) M3 subtype tends to have a good outlook while undifferentiated AML (MO) and acute megakaryoblast ic leukemia (M7) are harder to treat.o Chromosome changes: Children with leukemia cell translocations between chromosomes 15 and 17 or between 8 and 21, or with an inversion (rearrangement) of chromosome 16 have a better chance of being cured.o Response to treatment: Children whose leukemia responds quickly to treatment.o Body weight: Children within the normal weight range tend to do better than children who are underweight or overweight.

So the answer to this question is C


Instruction:
The method of presenting data of geographic distribution is:
A. Histogram
B. Pie chart
C. Frequency polygon
D. Cartogram

Response:
None

So the answer to this question is D


Instruction:
Chlorhexidine as a root canal irrigant is
A. Active against gram +ve , gram –ve bacterias and to some extent Virus with lipid envelop.
B. Normally used at 2% conc
C. A poly biguinide
D. All of the above

Response:
None

So the answer to this question is D


Instruction:
The smallest caranial nerve is the-
A. Olfactory
B. Oculomotor
C. Trochlear
D. Accessry

Response:
C i.e. Trochlear

So the answer to this question is C


Instruction:
In low doses, aspirin inhibits which of the following mediators?
A. Lipoxygenase
B. Cyclo-oxygenase
C. Thromboxane A2
D. Prostaglandin I2

Response:
Arachdonic acid metabolites are formed by two classes of enzymes such as Cycloxygenases and Lipoxygenases. Cycloxygenase pathway is mediated by COX1 and COX2 enzymes leading to the formation of Prostaglandins and Thromboxanes. In low doses, aspirin inhibit cycloxygenase irreversibly by acetylating one of its serine residues. Return of cycloxygenase activity depends upon synthesis of fresh enzyme. Ref: Essentials of Medical Pharmacology, 5th Edition, Pages 168, 173-4.

So the answer to this question is B


Instruction:
Transmission of hepatitis A virus occurs -
A. one week before and one week after onset of symptom
B. 2 weeks before onset of symptoms
C. 2 weeks after onset of symptoms
D. 1 week after onset of symptoms

Response:
The viruses shed in feces during the late incubation period and prodromal phase of illness REF:ANANTHANARAYANAN TTEXTBOOKOF MICROBIOLOGY 9EDITION PGNO.541

So the answer to this question is B


Instruction:
Investigation of choice in traumatic paraplegia is:
A. MRI
B. CT scan
C. Ultrasound
D. Beta rays

Response:
Ans. MRI

So the answer to this question is A


Instruction:
Subacute bacterial endocarditis is caused by
A. Streptococcus viridans
B. Haemolytic streptococci
C. Staphylococcus aureus
D. Nisseria

Response:
None

So the answer to this question is A


Instruction:
A 60 year old woman presents with generalized edema, skin ulceration and hypeension. Urine examination shows subnephrotic proteinuria (< 2gm) and microscopic haematuria. Serum complement levels are decreased and she is positive for anti- hepatitis c antibodies. The likely diagnosis is -
A. PSGN
B. Essential mixed cryoglobulinemia
C. Membrano proliferative glomerulonephritis
D. Focal segmental glomerulosclerosis

Response:
Essential mixed cryoglobulinemia is characterized by joint pains and swelling (ahritis), enlargement of the spleen, skin vasculitis with purplish patches, and nerve and kidney disease. This can lead to recurrent pain in the abdomen, hea attack, and bleeding in the lungs Essential mixed cryoglobulinemia is treated with combinations of medications which reduce inflammation and suppress the immune system. Medications used include nonsteroid antiinflammatory drugs (ibuprofen, aspirin, and others), steroids (prednisone, prednisolone), cyclophosphamide (Cytoxan), chlorambucil (Leukeran), and azathioprine (Imuran). Plasmapheresis (hemapheresis), a procedure whereby the blood's serum is filtered to remove the cryoglobulins, is also performed for severe symptoms. Ref Harrison20th edition pg 287

So the answer to this question is B


Instruction:
The after coming head of the baby in a breech delivery can be arrested by all of the following except:
A. Extended head
B. Hydrocephalus
C. Placenta pre
D. Incomplete dilatation of the cervix

Response:
If a whole baby can make its way from breech onwards till the head, then even if there is a placenta pre, it will not obstruct the fetal head to come though since there was no obstruction to the whole fetus.

So the answer to this question is C