INICET · Medical MCQs

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    • Anatomy
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    • Biochemistry
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    • Medicine
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Anatomy

3 questions
  1. Question 1 · Anatomy

    Which of the following lobes of the lung is affected when there is aspiration in supineposition?

    1. Lingula
    2. Upper part of right inferior lobe
    3. Right superior lobe
    4. Right middle lobe
    Show answer

    Answer: B. Upper part of right inferior lobe

    The upper part of the right inferior lobe is prone to aspiration in the supine position. It iscommonly seen in alcoholics. The posterior segment of the right superior lobe and the superior segment of the inferior lobes arethe most commonly involved lung sites when aspiration occurs in a recumbent patient. The location of aspiration depends on the position of the patient since it is gravity-dependent.Aspiration while standing leads to infiltration of bilateral inferior lobes because of larger andmore vertical corresponding bronchi, the middle lobes and the lingula. Aspiration of contents into foreign substances into the bronchial tree, usually oral or gastriccontents, produces aspiration pneumonia. Chemical (gastric acid) injury or bacterial infectioncomplicates the condition.

  2. Question 2 · Anatomy

    A professor underwent a left parotidectomy. Post surgery she complains of left lower lipparalysis. Which structure is involved?

    1. Facial nerve - Main trunk
    2. Facial nerve- Cervical branch
    3. Facial nerve-Temporal branch
    4. Parotid duct
    Show answer

    Answer: B. Facial nerve- Cervical branch

    Lower lip paralysis is caused by damage to the cervical branch of the facial nerve. The cervical branch of the facial nerve is responsible for innervating the platysma muscle, which isone of the muscles involved in the movements of the lower lip. Damage to this nerve canpotentially lead to paralysis of the lower lip. The facial nerve is the motor nerve of the face. Its five terminal branches emerge from the parotidgland and diverge to supply the various facial muscles as follows : • Temporal — frontalis, auricular muscles, orbicularis oculi • Zygomatic — orbicularis oculi (lower eyelid part) • Buccal — muscles of the cheek and upper lip • Marginal mandibular — muscles of the lower lip • Cervical — platysma

  3. Question 3 · Anatomy

    Which of the following is the first to be formed by the migration of the epiblast along theprimitive streak?

    1. Endoderm
    2. Cardiac mesoderm
    3. Epidermis
    4. Neural groove
    Show answer

    Answer: A. Endoderm

    Endoderm is the first to be formed by the migration of the epiblast along the primitive streak. Gastrulation is when the single-layered blastula reorganizes into a multi-layered structure calledthe gastrula. This process establishes the three primary germ layers that give rise to all tissues andorgans of the body. During gastrulation, epiblast cells migrate through the primitive streak toform the three germ layers: endoderm, mesoderm, and ectoderm. Cells from the epiblast migrate inward (invagination) through the primitive streak. The first group of migrating epiblast cells replaces the hypoblast to form the definitive endoderm.The endoderm gives rise to structures like the gastrointestinal tract, liver, and lungs. Subsequent migrating cells spread between the epiblast and the newly formed endoderm to createthe mesoderm, which forms structures like muscles, bones, and the cardiovascular system. Remaining epiblast cells that do not migrate through the primitive streak form the ectoderm,which gives rise to the epidermis and the nervous system. Other Options: Option B: Cardiac Mesoderm: The mesoderm forms after the endoderm and differentiates laterinto subtypes, including the cardiac mesoderm. Option C: Epidermis: The epidermis arises from the ectoderm, which forms after both theendoderm and mesoderm. Option D: Neural Groove: The neural groove is part of the ectoderm and forms duringneurulation, which occurs much later than the formation of the endoderm.

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Anesthesia

5 questions
  1. Question 4 · Anesthesia

    Ventilatory support is required if SpO2 becomes less than 90% because

    1. Tissues become vulnerable for hypoxia below 90%.
    2. Peripheral chemoreceptors becomes insensitive to hypoxia below 90%.
    3. Central chemoreceptors becomes insensitive to hypoxia below 90%.
    4. Patient capacity to generate adequate tidal volume decreases below 90%.
    Show answer

    Answer: A. Tissues become vulnerable for hypoxia below 90%.

    Tissues become vulnerable to hypoxia when SpO2 falls below 90% and thus supplementaryoxygen therapy is indicated below this oxygen saturation level. In adults, children, and infants (older than 1 month), When SpO2 falls below 90%, the PaO2becomes less than 60 mmHg and hence to prevent tissues from hypoxia, supplemental oxygentherapy should be started. In neonates, when SpO2 falls below 88%, the PaO2 becomes less than 50 mmmHg and hencesupplemental oxygen therapy should be initiated in these patients.

  2. Question 5 · Anesthesia

    How long should elective surgery be deferred for a patient who had cough and fever, after hehas recovered from COVID-19, and had not required any hospitalization?

    1. 6 weeks
    2. 8-10 weeks
    3. 12 weeks
    4. 4 weeks
    Show answer

    Answer: A. 6 weeks

    A patient, who has recovered from COVID-19 and hadn’t required any hospitalization should bedeferred from any elective surgery for a period of 6 weeks. The American Society of Anesthesiologists has suggested the following duration of intervals forpatients recovered from COVID-19. The wait time is from the date of diagnosis of COVID-19 to thedate of surgery as follows: • 4 weeks: for an asymptomatic patient or recovery from only mild, non-respiratory symptoms. • 6 weeks: for a symptomatic patient (e.g: cough, dyspnea) who did not require hospitalization • 8-10 weeks: for a symptomatic patient who is hospitalized or who has comorbidities such asdiabetes or who is immunocompromised. • 12 weeks: for a patient who was admitted in intensive care unit due to COVID-19 infection.

  3. Question 6 · Anesthesia

    Which of the following can be used for induction of anesthesia, maintenance and can besafely used in endotracheal intubation?

    1. Halothane
    2. Desflurane
    3. Sevoflurane
    4. Isoflurane
    Show answer

    Answer: C. Sevoflurane

    Among the given options, sevoflurane can be used for induction and maintenance of anesthesia. Itcan also be safely used in endotracheal intubation Inhalation agents such as sevoflurane and halothane are useful in the induction of anesthesia, Butsevoflurane due to its nonpungency and rapid onset makes it an excellent choice for smooth andrapid inhalation inductions in pediatric and adult patients. Sevoflurane has low blood solubilityand partition coefficients resulting in more rapid emergence. Sevoflurane also produces adequate muscle relaxation for intubation of children following aninhalation induction. Other options: Option A: Halothane can also be used for gaseous induction of anesthesia. Its disadvantage is longinduction time owing to the high blood:gas partition coefficient. However, the incidence ofhalothane hepatitis and myocardial depression has reduced its use. Option B: Desflurane produces laryngospasm when used for inhalational anesthesia and hence itis not used for induction. Laryngospasm is because of the pungent nature of the gas. Option D: Isoflurane causes respiratory depression as well as upper respiratory irritation.

  4. Question 7 · Anesthesia

    Ambu Aura 40 is an example of?

    1. Supraglottic laryngeal airway generation 1
    2. Supraglottic laryngeal airway generation 2
    3. Supraglottic laryngeal airway generation 3
    4. Supraglottic laryngeal airway generation 4
    Show answer

    Answer: A. Supraglottic laryngeal airway generation 1

    The Ambu Aura 40 is an example of a first-generation supraglottic airway device. Supraglottic airway (SGA) devices are classified into different generations based on their featuresand advancements: First-generation SGA devices: Basic devices designed to provide a seal around the glottis forventilation. They lack additional safety features, such as protection against aspiration. Examplesinclude the original Laryngeal Mask Airway (LMA Classic) and Ambu Aura 40 (can be reused upto 40 times), Second-generation SGA devices: These are improved devices with additional features to enhancesafety and functionality. They typically have: A gastric access channel (to allow gastric decompression or reduce the risk of regurgitation andaspiration). Improved sealing mechanisms. Examples: Ambu Aura Gain (with drain tube), LMA ProSeal, andi-gel. Third-generation and Fourth-generation devices: These are not standard classifications in clinicalpractice yet. However, devices beyond the second generation may feature advanced integrationwith monitoring systems, adjustable seals, or further innovations.

  5. Question 8 · Anesthesia

    Which of the following is incorrect regarding thiopentone?

    1. It is lipid soluble
    2. It achieves maximum concentration in the brain after an IV infusion
    3. Redistribution to adipose tissue and muscle is seen
    4. Rapid decline in its effect is due to renal excretion
    Show answer

    Answer: D. Rapid decline in its effect is due to renal excretion

    The rapid decline in the effect of thiopentone is due to its rapid redistribution, not due to renalexcretion. Thiopentone has a rapid onset of action because of its lipophilic nature, so it rapidly crosses theblood-brain barrier and enters highly perfused organs like the brain, heart, liver, and kidneys. As thiopentone redistributes from the brain to tissues: it perfuses to muscle, fat, and skin. Overtime, due to its high lipid solubility, thiopentone accumulates maximally in fat. Fat acts as astorage reservoir since it has low blood flow, and drug release from fat occurs slowly. This redistribution reduces the concentration of thiopentone, terminating its anesthetic effect. Other Options: Option A: It is lipid-soluble: Thiopentone is highly lipid soluble, which allows it to cross theblood-brain barrier quickly and achieve rapid onset of action in the CNS. Option B: It achieves maximum concentration in the brain after an IV infusion: Due to itslipophilicity, thiopentone rapidly accumulates in the brain after IV administration. Option C: Redistribution to adipose tissue and muscle is seen: Thiopentone’s effect is terminatedby redistribution from the brain to adipose tissue and muscle. This accounts for the short durationof its anesthetic effect.

Biochemistry

7 questions
  1. Question 9 · Biochemistry

    What is the effect of RNAi on gene expression?

    1. Knock out
    2. Knock down
    3. Knock up
    4. Knock in
    Show answer

    Answer: B. Knock down

    RNA interference(RNAi) is a biological process in which miRNA/siRNA molecules inhibit geneexpression ("knockdown") by neutralizing the targeted mRNAs. • miRNA regulates the expression of the target gene by post-transcriptional silencing. • Small interfering RNAs (siRNAs) are short RNA sequences that can be introducedexperimentally into cells and then serve as substrates for the Dicer enzyme and interact with theRNA-induced silencing complex (RISC) in a manner similar to that observed for endogenousmiRNAs. • Synthetic siRNAs targeted against specific mRNA species are useful in the study of gene functionand this technique is referred to as knockdown technology. • siRNAs are also being developed as possible therapeutic agents to silence pathogenic geneslike oncogenes involved in neoplastic transformation. Knock out: Mice generated through genetic manipulation of embryonic stem (ES) cells such that aspecific genetic locus is targeted and rendered non-functional ("knocked-out"), either by insertionof an irrelevant DNA sequence or by deletion of the endogenous DNA sequence. Knock-in: Mice generated through genetic manipulation of embryonic stem (ES) cells such that aspecific genetic locus is altered either by one-for-one substitution of DNA sequence or by additionof sequence not found in the endogenous genetic locus.

  2. Question 10 · Biochemistry

    What is the main substrate and metabolic fuel used by RBCs during fasting?

    1. Ketone bodies
    2. Glucose
    3. Free fatty acids
    4. Amino acids
    Show answer

    Answer: B. Glucose

    Glucose is the main substrate and metabolic fuel used by RBCs during fasting. RBCs lackmitochondria and are dependent fully on glycolysis (anaerobic) and pentose phosphate pathwayduring fasting state. Glucose is always required by brain and RBCs. Brain can meet only about 20 percent of energyrequirements by using ketone bodies and the remaining is supplied by glucose. Metabolic changesoccuring during fasting state and starvation therefore serve to preserve glucose and glycogenreserves for brain and RBCs.

  3. Question 11 · Biochemistry

    Banding technique used for dicentric chromosomes is:

    1. G banding
    2. C banding
    3. NOR
    4. R banding
    Show answer

    Answer: B. C banding

    The banding technique used for dicentric chromosomes is C banding. Heterochromatin includes non-coding regions of DNA, such as centromeric regions. The stainingof these constitutive heterochromatins, is done by C-banding using silver salt staining. A dicentricchromosome having two centromeres can hence be identified by C banding. Various techniques are used in karyotyping to study chromosomes. Staining with variouscompounds produces various results and helps us to understand the chromosomes better. Common techniques used are: • Q banding - Fluorescence staining with quinacrine mustard. • G banding - Giemsa staining after partial denaturation of chromatin. It is the most commontechnique used in cytogenetics to produce a visible karyotype by staining condensedchromosomes. • R banding - Reverse Giemsa staining. In this, the light and dark areas are reversed. • C banding - staining of heterochromatin with silver salts. • T banding - staining the telomeres of chromosomes.

  4. Question 12 · Biochemistry

    Misincorporation of uracil into DNA is seen in deficiency of:

    1. Vitamin B12
    2. Folate
    3. Vitamin B6
    4. Vitamin E
    Show answer

    Answer: B. Folate

    Misincorporation of uracil into DNA is seen in the deficiency of folate. Folate deficiency leads to decreased thymidylate synthesis from deoxyuridylate due to deficientmethylation of deoxyuridine monophosphate (dUMP) to thymidine monophosphate (TMP). Themethylation of dUMP to TMP is catalyzed by thymidylate synthase andrequires tetrahydrofolate (THFA) as a methyl group donor. This decreased TMP synthesiscan result in an imbalance of deoxyribonucleotides and hence leading to excessive uracilmisincorporation into DNA during replication and repair. Folic acid (vitamin B9) is a water-soluble vitamin, and its active form is tetrahydrofolate (THFA).It acts as a carrier of one-carbon groups. Tetrahydrofolate receives one-carbonfragments from donors such as serine, glycine, and histidine and transfers them to intermediatesin the synthesis of amino acids, purines, and pyrimidines. Folate deficiency results in: • Megaloblastic anemia • Homocystenemia • Neural tube defects (spina bifida, Anencephaly) in the fetus.

  5. Question 13 · Biochemistry

    Xanthurenic aciduria is seen in which of the following vitamin deficiency?

    1. Vitamin B2
    2. Vitamin B1
    3. Vitamin B12
    4. Vitamin B6
    Show answer

    Answer: D. Vitamin B6

    Xanthurenic aciduria is seen in deficiency of vitamin B6 (pyridoxal phosphate), a coenzymerequired by kynureninase enzyme to convert 3-OH kynurenine to 3-OH anthranilic acid duringtryptophan metabolism. Other options- Option A: Riboflavin or Vitamin B2 deficiency manifests as angular stomatitis, magenta tongue,seborrhea, corneal vascularization. It is seen in alcoholism, and individuals with a poor diet anddecreased intake of dairy products. Option B: A patient with thiamine or vitamin B1 deficiency presents either dryBeriberi with predominantly neurological symptoms like neuropathy, muscle weakness, andwasting or wet Beriberi with predominant cardiac symptoms like cardiomegaly, tachycardia, highoutput congestive cardiac failure. Option C: Vitamin B12 or cobalamin deficiency presents as anemia, fatigue, and pallor symptoms.Other presenting complaints may include peripheral neuropathy, glossitis, diarrhea, headaches,neuropsychiatric disturbances like dementia, and subacute combined degeneration of the spinalcord (SACD) leading to ataxia. A peripheral smear shows macrocytic anemia and hypersegmentedneutrophils.

  6. Question 14 · Biochemistry

    In a patient with Diabetes Mellitus, the liver produces more TG and VLDL. What is the causeof the same?

    1. Decreased action of LPL with increased action of HSL
    2. Decreased action of LPL with Decreased action of HSL
    3. Increased action of LPL with increased action of HSL
    4. Increased action of LPL with decreased action of HSL
    Show answer

    Answer: A. Decreased action of LPL with increased action of HSL

    In a patient with Diabetes Mellitus, the liver produces more TG and VLDL, due to decreasedaction of LPL with increased action of HSL. • Lipoprotein lipase (LPL): Enzyme responsible for hydrolyzing triglycerides (TG) in VLDL andchylomicrons into free fatty acids (FFAs) for storage or energy use in tissues. • Hormone-sensitive lipase (HSL): Enzyme that breaks down stored triglycerides in adipocytesinto free fatty acids and glycerol. Insulin has an inhibitory effect on hormone-sensitive lipase (HSL). In diabetes, low insulincorresponds to an increase in HSL activity. This promotes the breakdown of lipids and a release ofTG and free fatty acids from the adipose tissue. Insulin induces the activity of lipoprotein lipase (LPL) which is present on the capillary walls. LPLpromotes hydrolysis of TG and the degradation of the transport proteins, VLDL andchylomicrons. In diabetes, low insulin leads to decreased LPL activity which corresponds to anincrease in VLDL and TG. This results in increased production of triglycerides and VLDL in the liver.

  7. Question 15 · Biochemistry

    In the current scenario treatment for urea cycle disorder is available. Administration ofamino acid phenylbutyrate can decrease ammonia formation. Which amino acid providesphenylbutyrate?

    1. Arginine
    2. Glutamate
    3. Aspartate
    4. Glycine
    Show answer

    Answer: B. Glutamate

    The amino acid that provides phenylbutyrate in the treatment of urea cycle disorders is glutamate. Ammonia to Urea cycle: Ammonia gets converted to glutamate (non toxic form) throughtransamination. Glutamate is transported to the liver as glutamine (by glutamine synthase). Inthe liver, ammonia is released from glutamine, converting it to alpha keto glutatamte. Thereleased ammonia enters the urea cycle. In acetylation therapy, Phenylbutyrate (a prodrug) which is metabolized to phenylacetate.Phenylacetate combines with glutamine (a nitrogen-rich amino acid) to formphenylacetylglutamine. Phenylacetylglutamine is water-soluble and is excreted in the urine. Thisprocess helps trap excess nitrogen, bypassing the urea cycle. Other drugs that can be used to achieve the same outcome (ammonia scavengers) include: • Phenylacetate • Benzoate Other Options: Option A and C: Arginine and Aspartate do not play a role in this specific reaction. Option D: Glycine: The combination of benzoate and glycine forms hippuric acid, a non-toxiccompound. When Benzoate is administered to the body, it acts as a nitrogen-scavenging agent.Hippuric acid is water-soluble and is excreted in the urine, effectively removing one nitrogenmolecule from the body per molecule of hippuric acid formed. This mechanism is similar to howphenylacetate acts as a scavenger for glutamate.

Community Medicine

7 questions
  1. Question 16 · Community Medicine

    The most common mode of absorption of inorganic lead in industries leading to leadpoisoning is:

    1. Absorption through skin
    2. Inhaled lead dust
    3. Ingestion of contaminated food and water
    4. Contaminated hands
    Show answer

    Answer: B. Inhaled lead dust

    The most common mode of industrial lead poisoning (plumbism) is due to inhalation of fumesand dust of lead. Lesser common modes include ingestion of small quantities of lead trapped inthe upper respiratory tract, ingestion due to contaminated hands, and absorption through the skin(only organic lead). Lead poisoning is a notifiable and compensable disease in India. Lead exerts its toxic action by combining with the SH-group of enzymes involved in porphyrinsynthesis and carbohydrate metabolism, disrupting normal membrane permeability, and causingpotassium leakage in RBCs. Symptoms of lead toxicity include: Inorganic lead toxicity: • Abdominal colic, also known as Painter's colic • loss of appetite, obstipation, constipation • Anemia (stippling of red cells) • Wrist drop or foot drop • Blue-line on the gums known as Burtonian line Organic lead toxicity (mainly affects CNS): • Headache • Delirium, confusion • Insomnia Earliest and consistent sign: Facial pallor. Clinical symptoms of lead poisoning occur when blood levels are >70µg/dl. Measurement ofCoproporphyrin (CPU) in urine is a useful screening test. Periodic examination of workers mustbe done and atmosphere lead concentration must be kept <2mg/cubic meter at the workspace.Handwashing before eating and prohibition of taking food in workplaces is essential.D-penicillamine is used as a chelating agent for treatment.

  2. Question 17 · Community Medicine

    According to WHO, which of the following are components of the Human Development Index(HDI)?

    1. Life expectancy at birth, literacy and standard of living
    2. Life expectancy at 1 year, literacy and standard of living
    3. Infant mortality rate, literacy and standard of living
    4. Life expectancy at 1 year, literacy and infant mortality rate
    Show answer

    Answer: A. Life expectancy at birth, literacy and standard of living

    The components of the Human Development Index (HDI) include life expectancy at birth,literacy, and standard of living. Human Development Index (HDI) is a measure of a country's average achievements in threedimensions of human development: • A long and healthy life, as measured by life expectancy at birth • Knowledge (literacy), as measured by mean years of schooling and expected years of schooling,and • A decent standard of living, as measured by GNI (gross national income) per capita. HDI values range from 0 to 1, with 1 as the maximum value. Note- Life expectancy at 1 year, literacy and infant mortality rate are components of the Physicalquality of life index (PQLI).

  3. Question 18 · Community Medicine

    Which of the following is true about the management of a baby born to an HIV-positivemother?

    1. If replacement feeding, the neonate should be tested for HIV, if negative, ART is not required.
    2. In exclusive breastfeeding, give nevirapine for 6 months only when baby is positive
    3. If replacement feeding, the neonate need not receive nevirapine if HIV negative
    4. If exclusive replacement feeding, give nevirapine for 6 weeks, even if the baby is negative
    Show answer

    Answer: D. If exclusive replacement feeding, give nevirapine for 6 weeks, even if the baby is negative

    If exclusive replacement feeding, giving nevirapine for 6 weeks, even if the baby is negative is trueregarding the management of a baby born to an HIV-positive mother. The national AIDS control organization (NACO) has recommended the following for an infantborn to an HIV-positive mother. According to the National AIDS Control Organization (NACO), antiretroviral therapy (ART)should be given to all HIV-positive pregnant women including during labor hours andbreastfeeding. They are started on a triple-drug ART. ART should be given irrespective of the CD4count, and clinical stage and should be continued lifelong. Immediate care of HIV-exposed infants includes wiping of baby's mouth and nostrils, handling ofinfants with gloves, immediate cord clamping, and initiation of feeding. An infant born to anHIV-positive mother should be breastfed exclusively or by replacement feeds exclusively for 6months of age. Infants should receive ARV prophylaxis. A single drug or dual drug regimen is followed dependingon the risk of transmission which can be low or high risk. • Low-risk infants are those born to mothers with a low viral load which is detected after 32 weeksof pregnancy till delivery. • High-risk infants are those born to HIV-positive mothers, not on ART, with viral load notestimated beyond 32 weeks, viral load not suppressed after 32 weeks, or diagnosed with HIVwithin 6 weeks of delivery. HIV RiskARV prophylaxisDuration Low-risk infantsSyrup nevirapine or zidovudineBirth to6 weeksof ageirrespective of feeding High-risk infantsSyrup nevirapine + syrup zidovudineExclusive replacement feeding: Birth to 6 weeksExclusive breastfeeding: Birth till 12 weeks

  4. Question 19 · Community Medicine

    Contraceptive failure is assessed using which of the following methods?

    1. Pharmacological index
    2. Pearl index
    3. Performance index
    4. Efficacy index
    Show answer

    Answer: B. Pearl index

    Pearl index is the common method used to measure contraceptive failure rate. It is defined as the number of failures per 100 woman-years of exposure (HWY). This rate is givenby the formula: Pearl index = Total accidental pregnancies x 1200/Total months of exposure

  5. Question 20 · Community Medicine

    High fat intake according to FDA is?

    1. Diet should contain 400 kcal from carbohydrates
    2. Diet should contain total 1000 kcal
    3. Diet should contain 800 kcal from fat
    4. Diet should contain 80% fats
    Show answer

    Answer: B. Diet should contain total 1000 kcal

    High fat intake according to FDA is a diet that should contain a total of 1000 kcal. High fat diets have low carbohydrates, <50 grams/day. This equates to <250 kcal/day in theform of carbohydrates, not 400 kcal (option A). The percentage of calories from fat is >50%(option D), which would equate to about 500-600 kcal from fats (option C). The diet shouldcontain a total of 800 - 1000 kcal. Ketogenic diets mainly incorporate low carbohydrates (<50 grams/day), low protein, and highfats. Gluconeogenesis from the glucogenic amino acids in the diet and fat oxidation are utilized tomeet the metabolic demands of the body. Ketogenic diets are especially useful for the treatment ofintractable epilepsy.

  6. Question 21 · Community Medicine

    A mother brought her 6 month old infant baby to an Anganwadi worker Lata for adviceregarding the vaccination of her child. The last vaccines he had received were OPV3, PENTA3, IPV2, and ROTA3 at 14 weeks of age. BCG and Hep B were given at birth. When is the nextvaccination due?

    1. MR vaccine at left arm intramuscularly and Vitamin A drops at the end of 9 months
    2. MR vaccine at right arm subcutaneously and Vitamin A drops at the end of 8 months
    3. MR vaccine at left arm intramuscularly and Vitamin A drops at the end of 8 months
    4. MR vaccine at right arm subcutaneously and Vitamin A drops at the end of 9 months
    Show answer

    Answer: D. MR vaccine at right arm subcutaneously and Vitamin A drops at the end of 9 months

    According to the National Immunization Schedule, the Measles/MR vaccine is given at 9completed months (1st dose), and a second dose is given at 16-24 months. Measles vaccine can begiven till 5 years of age. A dose of 0.5 mL is administered subcutaneously in the right upper arm. Vitamin A schedule in the National Immunisation Schedule(NIS) • 1 lakh IU at 9 completed months along with the Measles-Rubella vaccine • 2 lakh IU is administered bi-annually for children between 1-5 years of age Total number of doses administered- 9 doses,17 lakhs IU

  7. Question 22 · Community Medicine

    Which of the following is not a polysaccharide vaccine?

    1. Pneumococcal
    2. Meningococcal
    3. Hepatitis B
    4. Haemophilus influenza type B
    Show answer

    Answer: C. Hepatitis B

    Hepatitis B vaccine is a recombinant protein vaccine. Polysaccharide vaccines are a type of subunit vaccine containing the capsular polysaccharideantigen. They generate a T-cell independent response. Examples include: • Pneumococcus • Meningococcus • Haemophilus influenzae type b (Hib) Recombinant protein vaccines: With the help of recombinant DNA technology, the purifiedrecombinant proteins from the cultured microorganisms result in cleaner vaccine preparationswith a better safety profile. Examples include: • HBV • HPV • Lyme disease • Cholera toxin-B • COVID-19

ENT

3 questions
  1. Question 23 · ENT

    A 35-year-old female patient presents with complaints of nasal obstruction and post-nasaldrip. There is a past history of FESS for failed conservative management 5 years ago.Uncinectomy and maxillary ostium dilation was done during the previous FESS. A DNE donenow shows patent ostia and mucosal edema of the maxillary sinus lining. What is the nextbest step in management?

    1. Repeat surgery
    2. Macrolides for 4 months
    3. Steroid irrigation and antihistamines
    4. Biological therapy
    Show answer

    Answer: C. Steroid irrigation and antihistamines

    The history of nasal obstruction and post-nasal drip and endoscopy showing mucosal edema issuggestive of acute rhinosinusitis. The next best step in management is symptomatic treatmentwith nasal steroid irrigation, decongestants, and antihistamines, regardless of the patient's pasthistory of FESS operation. Acute viral rhinosinusitis causes mucosal edema which causes obstruction of the sinus ostia.There is also reduced mucociliary transport in the sinuses resulting in stasis of secretions withinthe sinuses causing headache, facial pain, congestion, nasal obstruction or discharge, post-nasaldrip, and hyposmia or anosmia. If the sinuses remain obstructed or the mucociliary transport system does not return to normal, abacterial infection can ensue, causing fever, purulent discharge, and persistent or worseningsymptoms beyond 10 days. In the case of bacterial sinusitis, antibiotics like amoxicillin and macrolides (erythromycin) areused for a short course (Option B). Procedures like antral lavage and surgery (Option A) are done in cases not responding to medicaltherapy. Biological therapy (Option D) is not used in the treatment of sinusitis.

  2. Question 24 · ENT

    What test is not used for testing olfaction?

    1. Smell diskettes
    2. Arnold stick test
    3. UPSIT
    4. CC-SIT
    Show answer

    Answer: B. Arnold stick test

    Arnold stick test is not a test used for olfaction. Tests used for olfaction are as follows: • Smell diskettes are used as a screening test for olfaction. Reusable diskettes are used with theapplication of 8 different odorants. A questionnaire with illustrations is used to score the patient. • The University of Pennsylvania smell identification test (UPSIT) is a commercially available,self-administered test used to assess the function of an individual's olfactory system via smellidentification. It uses microencapsulated odorants, which are released on scratching standardizedodor-impregnated test booklets. • The 12-item version of UPSIT, also known as the cross-cultural smell identificationtest (CC-SIT), is useful for quantifying smell loss when less time is available. This test serves as ameans to reliably assess olfactory function in less than 5 minutes. • The Sniffin' sticks test is a new test of nasal chemosensory performance based on pen-likeodor-dispensing devices.

  3. Question 25 · ENT

    Which of the following is the most dangerous lesion?

    1. U/L Partial RLN lesion
    2. B/L Partial RLN lesion
    3. U/L Complete RLN lesion
    4. B/L Complete RLN lesion
    Show answer

    Answer: D. B/L Complete RLN lesion

    Bilateral complete RLN (recurrent laryngeal nerve) lesion is the most dangerous lesion among thegiven options. In bilateral RLN paralysis • Both the cords lie in the median or paramedian position, the airway is inadequate causingdyspnea and stridor. This may lead to death if untreated. • The voice is good. • Dyspnea and stridor become worse on exertion or during an attack of acute laryngitis. All laryngeal muscles are supplied by the ipsilateral recurrent laryngeal nerve (RLN), exceptcricothyroid which is supplied by the superior laryngeal nerve (SLN). Both SLN and RLN arebranches of the vagus nerve. The recurrent laryngeal nerves (RLN) on the right and the left side follow different courses- • Right RLN takes origin in the root of the neck, winds around the first part of the subclavianartery, and then ascends in the tracheoesophageal groove. • Left RLN takes origin in the thorax, winds around the arch of the aorta, and then ascends in thetracheoesophageal groove. Since it has a longer course, it is more likely to be injured.

Forensic Medicine

7 questions
  1. Question 26 · Forensic Medicine

    Sagar was being tried in a court for the murder of Mohan in the park. Shikhar appeared incourt and said he saw Sagar carrying a knife in his hand while in the park. What is thisevidence known as?

    1. Direct
    2. Indirect
    3. Hostile
    4. Hearsay
    Show answer

    Answer: B. Indirect

    This evidence is known as indirect evidence. Indirect evidence is also known as circumstantialevidence. It is not the direct testimony of an eyewitness. It has an effect on other facts andsubsidiary facts which are depended upon as consistent. Other options: Option A: Direct evidence is the evidence of a fact that is actually in issue. Eg: The witness saw Ashooting B, prescription, consent form Option C: Hostile evidence is given by someone who is supposed to have some interest or motivefor hiding part of the truth. He might give completely false evidence too. Option D: Hearsay evidence is given by someone about something that he did not personallywitness. Sometimes it might be obtained from a third party. It is presented in the court toclaim that the facts contained in the statement are true.

  2. Question 27 · Forensic Medicine

    Which of the following is not a feature of cocaine intoxication?

    1. Tingling and numbness of hands and feet
    2. Constricted pupils and bradycardia
    3. Hypertension
    4. Hyperthermia
    Show answer

    Answer: B. Constricted pupils and bradycardia

    Constricted pupils and bradycardia are not a feature of cocaine intoxication. Dilated pupils andrapid pulse are seen in cocaine intoxication. Cocaine is obtained from the leaves of Erythroxylum coca. Cocaine can beadministered by snorting, intravenous route, or smoking. The manifestations of cocaine intoxication occur in two stages- the stage of excitement and thestage of depression. In the stage of excitement, the features seen are dry mouth, dysphagia, tachycardia, tachypnea,hypertension, dilated pupils, and increased temperature. There may be a feeling of well-being, andthe patient may be excited, restless, and talkative. This progresses into a calm, dullstate. Neurological manifestations include exaggerated reflexes, tremors or convulsions,hallucinations, tingling, and numbness in the hands and feet. The stage of depression is characterized by feeble respiration, excessive sweating, collapse,convulsions, and death. The fatal dose of cocaine is 1g when taken orally. The fatal period ranges from a few minutes tohours. Chronic cocaine users may have the feeling of insects crawling on their skin or the presence ofgrains of sand under the skin. This is known as cocaine bugs or Magnan's symptom. Ulceration ofthe nasal septum may be seen in chronic abusers who sniff cocaine. Treatment includes gastric lavage with warm water containing potassium permanganate,charcoal, or tannic acid if taken orally. When it is through the nose or throat, a wash with water isdone. In case of intravenous intoxication, a ligature is applied above the site of the injection. The antidote for cocaine intoxication is inhalational amyl nitrite. Other measures include controlof convulsions, airway and circulatory stabilization, administration of iv thiamine, and naloxone.

  3. Question 28 · Forensic Medicine

    Treatment of choice for acute arsenic poisoning is :

    1. Ipecac
    2. Penicillamine
    3. Activated charcoal
    4. Dimercaprol
    Show answer

    Answer: D. Dimercaprol

    Dimercaprol is the treatment of choice for acute arsenic poisoning. Arsenic is an odorless, tasteless, metallic poison. Gastrointestinal features of acute arsenicpoisoning include nausea and vomiting, colicky abdominal pain, and profuse diarrhea with ricewater stools. In chronic arsenic poisoning, skin symptoms include erythematous lesions, and flushing and canalso lead to hyperkeratotic skin and desquamation. Pigmentation is patchy andresembles raindrops on a dusty road. In the distal parts of the body, hyperkeratosis ispresent. Mees lines are transverse white striae of the fingernails, called striate leukonychia. Renalmanifestations include chronic nephritis and urine contains albumin, casts, etc. Arsenic can be detected in the bones for prolonged periods. Hence it can be found evenin skeletonized remains of the body. Other options: Option A: Ipecac is an emetic used in ergot and hydrocarbon poisoning. Option B: Penicillamine is used in lead poisoning as a chelating agent. Option C: Activated charcoal acts as an adsorbing agent in many poisonings like aspirin,phenobarbital poisoning, etc.

  4. Question 29 · Forensic Medicine

    In the 18th century, the British East India Company smuggled which of the following drugs toChina?

    1. Cocaine
    2. Magic mushroom
    3. Opium
    4. Marijuana
    Show answer

    Answer: C. Opium

    The British East India Company smuggled opium into China in exchange for tea, silk andporcelain pots. Opium derived from the immature seedpods of the opium poppy (Papaver somniferum), wasoriginally indigenous to the present-day territory of Turkey. The introduction of opium to Chinaoccurred during the late 6th or early 7th century CE, facilitated by Turkish and Arab traders. During this time, tobacco smoking moved from North America to China, resulting in widespreadopium smoking. Opium addiction and imports increased in the first century of the Qing era(1644–1911/12). To address the issue, the Yongzheng emperor (1722–35) banned opium traffickingand consumption in 1729. Despite hurdles, the trade continued, and in 1796, the Jiaqing emperorbanned opium imports and cultivation. Opium sales survived despite these laws. In the early 18th century, the Portuguese made a lot of money importing opium from India andselling it in China. British traders became China's main suppliers in 1773. The British East IndiaCompany monopolized Bengali opium farming. The company developed an inexpensive andproductive opium poppies cultivation method in this location. The US and other Western nationstraded Turkish and Indian opium. The opium trade was pursued by Britain and other European nations as a means to address theirpersistent trade deficit with China. In Europe, there existed a substantial and notable demand forChinese commodities such as tea, silks, and porcelain pots. Conversely, in China, there was a verylimited market for Europe's manufactured goods and other trade items. As a result, Europeanswere required to utilize gold or silver as payment for Chinese goods. The opium trade, whichresulted in a consistent demand for opium among Chinese individuals addicted to the substance,effectively addressed the persistent trade imbalance by facilitating the importation of opium fromWestern countries. Opium was not carried by the East India Company. Due to the Chinese restriction, thecorporation outsourced it to “country traders”—private traders entitled to transport commoditiesfrom India to China. Smugglers on the Chinese shore bought opium from country traders. Thetraders gave the East India Company their sales of gold and silver. The corporation spent its goldand silver in China on profitable things to sell in England. Opium imports into China rose from 200 chests in 1729 to 1,000 in 1767 and 10,000 between1820 and 1830. Although chest weight varied by origin, it averaged 140 pounds (63.5 kg). By 1838,China imported 40,000 chests annually. China lost the balance of payments to Britain for the firsttime. Opium distribution networks had evolved across China, frequently with corrupt officials'help. High opium addiction levels affected imperial troops and officials.

  5. Question 30 · Forensic Medicine

    In a Firearm injury, blackening surrounding the gunshot entry wound is due to?

    1. Flame effect
    2. Unburnt powder
    3. Smoke deposition
    4. Muzzle of the firearm
    Show answer

    Answer: C. Smoke deposition

    Blackening occurs due to the deposits of smoke over the epidermis which can be removed with awet cloth. This can be seen up to around 15 cm of firing. Other injuries related to firearms are: • Burn injuries - occur due to flame released during the firing. It is seen in contact shot wounds ordistances up to 5-8 cm. Singeing of hair is caused due to burns. • Tattooing - occurs due to the deposition of unburnt powder particles. These are difficult toremove. They are also called stippling or peppering. This is seen up to 50 cm of firing • Greasing and collaring - occur due to the deposition of the lubricant of the bullet in the tissues,during its penetration. It can be found in contact, near-shot, or distant shot wounds. In a firm contact shot, the imprint of the muzzle of the gun is seen on the skin around the wound.The victim is within the range of flame, hence singeing of the hair will be present. Tattooingbegins only when the muzzle-to-target distance is >1cm, like in a loose contact shot. When the distance between the victim and the firing is 5-8cm, it is called a close shot. The victimis within the range of the flame, hence, features related to flame, smoke, and powder depositionwill be seen. Singeing, burning, blackening and tattooing will be present. When the distance between the victim and the firing is >50 cm, it is a distant shot. The victimlies outside the range of flame and powder deposition. Hence, features related to flame, smoke,and powder deposition are absent. Burning, blackening, and tattooing are absent.

  6. Question 31 · Forensic Medicine

    What is the Declaration of Oslo related to?

    1. Human Torture
    2. Therapeutic abortion
    3. Human experimentation
    4. Hunger strikes
    Show answer

    Answer: B. Therapeutic abortion

    In 1970, the World Medical Association adopted a resolution on therapeutic abortion known asthe Declaration of Oslo. Other options: Option A: Declaration of Tokyo: related to human torture and other degradation Option C: Declaration of Helsinki: concerned with human experimentation. Options D: Declaration of Malta: related to hunger strikes.

  7. Question 32 · Forensic Medicine

    Marshall's triad (punctate lacerations, bruises, and abrasions) is seen in which type of blastinjury?

    1. Primary blast injury
    2. Secondary blast injury
    3. Tertiary blast injury
    4. Quaternary blast injury
    Show answer

    Answer: B. Secondary blast injury

    Marshall's triad in a bomb explosion is seen due to secondary blast injuries. Marshall described the characteristic injuries associated with bomb explosions. It includes: • Abrasions • Bruises/ contusions • Puncture lacerations Blast injuries are classified as: • Primary injuries- effect of transmitted blast waves on gas-containing structures. External injuriesare absent and are characterized by internal injuries. The middle ear, lungs, and gastrointestinaltract are the most susceptible to this. • Secondary injuries- injuries caused by shrapnel and other airborne debris propelled by theexplosion. These injuries may affect any part of the body and sometimes result in penetratingtrauma. Most casualties are caused by secondary injuries. • Tertiary injuries- are caused by blast wind that can throw victims against solid objects. • Quaternary injuries- all other injuries caused by explosions are not included in the above threecategories. eg, flash burns, crush injuries, respiratory effects due to toxic dust, gas, etc.

Medicine

7 questions
  1. Question 33 · Medicine

    Which of the following is not a major diagnostic criterion of neuromyelitis optica?

    1. Area postrema syndrome
    2. Acute Myelitis
    3. Focal epilepsy
    4. Optic neuritis
    Show answer

    Answer: C. Focal epilepsy

    Focal epilepsy is not considered a major diagnostic criterion of neuromyelitis optica (NMO). Neuromyelitis optica (Devic’s disease) is an aggressive inflammatory demyelinating disordercharacterized by recurrent attacks of optic neuritis and myelitis. It is associated with a highly specific autoantibody directed against aquaporin-4(AQP-4). AQP-4 islocalized to the foot processes of astrocytes in close proximity to endothelial surfaces. The core clinical characteristics include optic neuritis, acute myelitis, area postrema syndrome(unexplained hiccups/ nausea/ vomiting), acute brainstem syndrome, symptomatic narcolepsy oracute diencephalic clinical syndrome with Neuromyelitis Optica Spectrum Disorder(NMOSD)-typical diencephalic MRI lesions, and symptomatic cerebral syndrome withNMSOD-typical brain lesions. Attacks of optic neuritis are bilateral and severe in NMO, whereas in multiple sclerosis (MS) opticneuritis is usually unilateral. Progressive symptoms are not typical of NMO, unlike in MS. CSFfindings in NMO show higher pleocytosis, with neutrophils and eosinophils in acute cases, andrare occurrence of oligoclonal bands, distinguishing it from MS. Acute attacks of NMO are treated with high-dose glucocorticoids. Plasma exchange can be used inacute attacks that do not respond to glucocorticoids. Drugs used for prophylaxis against relapses include mycophenolate mofetil, rituximab,glucocorticoids, and azathioprine. Three monoclonal antibodies approved for attack preventioninclude eculizumab (terminal complement inhibitor), inebilizumab (B-cell depleter), andsatralizumab (IL-6 receptor blocker). NMO typically presents as a recurrent disease, with a monophasic course observed in less than10% of patients. Untreated NMO can lead to disability, including respiratory failure, blindness,and limb paralysis.

  2. Question 34 · Medicine

    For the diagnosis of resistant hypertension, the patient should have uncontrolledhypertension despite taking three antihypertensive drugs. This should include at least one:

    1. Calcium channel blocker
    2. Beta blocker
    3. Alpha blocker
    4. Diuretic
    Show answer

    Answer: D. Diuretic

    For the diagnosis of resistant hypertension, the patient should have uncontrolled hypertensiondespite taking three antihypertensive drugs, including at least one diuretic. While a calcium channel blocker, beta blocker, or alpha-blocker can be part of theantihypertensive regimen, the inclusion of a diuretic is necessary for the diagnosis of resistanthypertension. The diuretic helps in addressing fluid retention which is often a significantcontributor to hypertension. Resistant hypertension is more common in patients aged >60 years than in younger patients. Itmay be related to nonadherence to therapy, obesity, and excessive alcohol intake. In the absence of a specific identifiable cause, mineralocorticoid receptor antagonists likespironolactone have been found to be most effective add-on drugs for the treatment of resistanthypertension.

  3. Question 35 · Medicine

    Which of the following is the most characteristic feature of Kallmann syndrome?

    1. Anosmia
    2. Syndactyly in males
    3. Precocious puberty in females
    4. White forelock
    Show answer

    Answer: A. Anosmia

    Anosmia is the most characteristic feature of Kallman's syndrome. Kallman's syndrome is a hypothalamic disorder that occurs due to defectivegonadotropin-releasing hormone (GnRH). It is mainly associated with anosmia or hyposmia or hypoplasia. Other associated features such ascolour blindness, optic atrophy, deafness, cleft palate, renal abnormalities, and cryptorchidismmay be present. The presenting features in males are delayed puberty and micropenis while females present withprimary amenorrhoea and failure of secondary sexual development. The condition is characterised by low luteinizing hormone (LH), follicle-stimulating hormone(FSH), testosterone and estradiol. Treatment includes human chorionic gonadotropin (hCG) or testosterone in males for restorationof puberty and secondary sexual characteristics. In females, cyclic estrogen and progestin can begiven. Fertility is restored by giving gonadotropins or pulsatile GnRH

  4. Question 36 · Medicine

    Which of the following does not help in the localization of lesions in the spinal cord?

    1. Contralateral hemiplegia
    2. Fasciculation at the level of lesion
    3. Upper motor neuron lesion and lower motor neuron lesion
    4. Bladder involvement
    Show answer

    Answer: A. Contralateral hemiplegia

    Contralateral hemiplegia does not help in the localization of the lesion in the spinal cord. Contralateral hemiplegia is a feature of involvement of the corticospinal tract from its origin in thecerebrum (motor homunculus) till it crosses at the level of the lower medulla. At the level of spinalcord lesion, it leads to ipsilateral hemiplegia (eg: Brown Sequard syndrome) Fasciculation at the level of lesion, upper motor neuron lesions (UMNL) and lower motor neuronlesions (LMNL), and bladder involvement are features of spinal cord lesions. The presence of fasciculation (option B) indicates segmental signs corresponding to the disturbedindividual cord segment. Other features include altered sensation, muscle atrophy, or absent deeptendon reflex. Upper motor neuron (UMN) lesions and lower motor neuron (LMN) lesions (option C) can helplocalize lesions in the spinal cord. UMN signs include paraplegia or quadriplegia with exaggerateddeep tendon reflexes, positive Babinski sign, and spasticity. LMN lesions include muscle atrophy,fasciculations, decreased reflexes, and flaccid paralysis. Autonomic disturbances such as absent sweating below the implicated cord level and bladderinvolvement (option D) are seen when there is transverse damage to the spinal cord.

  5. Question 37 · Medicine

    During which phase of the cardiac cycle is an AR murmur likely to be heard?

    1. Ventricular ejection
    2. Throughout cardiac cycle
    3. Isovolumetric contraction
    4. Ventricular diastole
    Show answer

    Answer: D. Ventricular diastole

    Aortic regurgitation (AR) murmur is most likely to be heard during the ventricular diastole. On auscultation of patients with chronic AR, a high-pitched decrescendo diastolic murmur isheard. It can be best noted in the third intercostal space on the left sternal border. In patientswith mild AR, the murmur is brief. But as the disease progresses, it becomes longer in duration,and can indeed become holodiastolic. In patients in whom AR is due to primary valvular disease,the diastolic murmur is heard louder on the left sternal border. However, if the murmur is louderon the right sternal border, it indicates that AR is due to aneurysmal dilation of the aortic root.The auscultatory features of AR are intensified by handgrip, which augments systemic vascularresistance and increases LV afterload. The heart murmurs heard in systole are: • Aortic and pulmonic stenosis: ejection systolic murmur • Chronic mitral and tricuspid regurgitation: holosystolic murmur • Mitral valve prolapse: midsystolic click followed by a late systolic murmur • Ventricular septal defect: holosystolic murmur The heart murmurs heard in diastole are: • Aortic regurgitation: early diastolic, decrescendo murmur • Mitral stenosis: opening snap following by mid to late diastolic murmur

  6. Question 38 · Medicine

    In which of the following malignancies, a liver transplantation is not contraindicated?

    1. Renal carcinoma
    2. Breast carcinoma
    3. Melanoma
    4. Testicular cancer
    Show answer

    Answer: D. Testicular cancer

    Liver transplantation is relatively not contraindicated in testicular cancer. Past history ofmalignancy is a relative contraindication to transplantation, because the effect of surgery andresulting immunosuppression may predispose the patient to early recurrence of the tumor. Testicular carcinoma has a relatively low risk of recurrence (0%-10%) following transplantation.In contrast, renal carcinoma, breast carcinoma, and melanoma pose a higher risk of recurrence,exceeding 10%.

  7. Question 39 · Medicine

    Which is not used as a diagnostic criterion for diabetic ketoacidosis?

    1. Serum glucose>250mg/dl
    2. Serum potassium <3.5mEq/L
    3. Serum bicarbonate <15mEq/L
    4. ph <7.3
    Show answer

    Answer: B. Serum potassium <3.5mEq/L

    Serum potassium <3.5mEq/L is not a diagnostic criterion of diabetic ketoacidosis. DKA is an acute complication of type 1 diabetes mellitus. It is usually precipitated by absolute orrelative insulin deficiency caused by increased metabolic demand (infections) or insulinnoncompliance. The common symptoms include nausea, vomiting, and severe abdominal pain. The common signsinclude Kussmaul breathing, fruity odor of the breath, dehydration, tachycardia, hypotension, anddelirium. The diagnosis of DKA is made based on the following parameters: • Hyperglycemia > 250 mg/dl • Ketosis or ketones in urine • Metabolic acidosis with pH < 7.3 • HCO3 < 15-18 mEq/L • High anion gap due to ketoacids and lactic acidosis Hyponatremia is seen as a result of hyperglycemia (osmotic shift causes dilutional hyponatremia).Normal sodium levels on admission indicate severe dehydration. Potassium levels at presentationmay be normal or elevated despite total body deficit (due to intracellular shift).

Microbiology

7 questions
  1. Question 40 · Microbiology

    Which cell is involved in type 4/delayed type hypersensitivity?

    1. Cytotoxic T cell
    2. Macrophage
    3. Dendritic cell
    4. Neutrophil
    Show answer

    Answer: A. Cytotoxic T cell

    Cytotoxic T cells are involved in type 4/delayed type hypersensitivity

  2. Question 41 · Microbiology

    A 24-year-old male patient presented to the outpatient department with cervicallymphadenopathy. As a part of the evaluation, the Paul-Bunnell test was done. What is theprinciple of this test?

    1. Complement fixation
    2. Super-antigens
    3. Heterophile antibodies
    4. Haptens
    Show answer

    Answer: C. Heterophile antibodies

    The principle of the Paul-Bunnell test is based on heterophile antibodies. Heterophile antibodieshave the ability to agglutinate shared antigens. Paul-Bunnell test is the standard diagnostic procedure for infectious mononucleosis. In infectiousmononucleosis, heterophile antibodies agglutinate sheep erythrocytes. However, such antibodiesmay also occur after injections of sera and sometimes even in normal individuals. Infectiousmononucleosis antibodies may be differentiated by absorption tests. An agglutination titer of 100or above is suggestive of infectious mononucleosis. The Weil-Felix reaction is also a heterophile agglutination test. It is based on the sharing of acommon antigen between typhus rickettsiae and some strains of proteus bacilli. It is used for theserodiagnosis of typhus fevers. Other examples based on the same principle are Streptococcus MGagglutination test for the diagnosis of primary atypical pneumonia and the cold agglutination testfor diagnosis of mycoplasmal (primary atypical) pneumonia. Other options: Option A: Complement fixation test is based on the ability of antigen-antibody complexes tofix complement. This is a very sensitive test capable of detecting various types of antigens andantibodies. Option B: Superantigens are certain protein molecules, that activate very large numbers of T cellsirrespective of their antigenic specificity such as staphylococcal enterotoxins. Superantigens bindoutside the antibody binding groove directly to the lateral aspect of the TCR beta chain. Microbialsuperantigens are medium-sized proteins characterized by high resistance to proteases and todenaturation by CD4+ T cells. They cause the release of cytokines (IL-2) which results in massiveproliferation of T lymphocytes. This ultimately leads to further release of a variety of cytokineswhich can have profound effects on the immune system. Option D: Haptens are substances that can react specifically with antibodies but are incapable ofinducing antibody formation alone. Haptens become immunogenic (capable of inducingantibodies) by combining with a larger molecule carrier. They may be complex or simple. Whilecomplex haptens can precipitate with specific antibodies, simple haptens are non-precipitating.

  3. Question 42 · Microbiology

    Which of the following is wrong regarding primary and secondary immune response ?

    1. 1
    2. 2
    3. 3
    4. 4
    Show answer

    Answer: D. 4

    Statement 4 is wrong regarding primary and secondary immune responses. Primary immuneresponse has lower antibody affinity and vice versa. An immune response is a specific reactivity that occurred in a host as a result of an antigenicstimulus. It is of two types: Humoral or antibody-mediated response and cellular or cell-mediatedresponse. The response to the antigenic stimulus can be primary or secondary. The features of the primaryand secondary immune responses are as follows FeaturesPrimary immune responseSecondary immune response 1. Lag period4-10 days1-3 days 2. Peak response time7-10 days3-5 days 3. Antibody responseDepends on the antigenMore potent than the primary immune response 4. Isotype generateIgMIgG 5. AntigensBoth thymus-dependent and independentDepends only on the thymus 6. Antibody affinityLowerHigher

  4. Question 43 · Microbiology

    Which of the following is an example of Mollicutes?

    1. Ureaplasma urealyticum
    2. Orientia tsutsugamushi
    3. Chlamydophila pneumoniae
    4. Chlamydia trachomatis
    Show answer

    Answer: A. Ureaplasma urealyticum

    Ureaplasma urealyticum belong to the class Mollicutes which are also known as cell-wall-freebacteria. This class contains around 200 species of mycoplasma. Mycoplasma are the smallest microorganisms that can thrive in a cell-free medium. They arepleomorphic and lack cell walls. There are 4 species that are of importance in humans. • Ureaplasma urealyticum • Mycoplasma pneumoniae • Mycoplasma hominis • Mycoplasma genitalium It can be grown in fluid or solid media. Colonies typically have a fried egg appearance. Theymainly cause pneumonia and genital infections. Ureaplasma urealyticum is unique in it's ability tohydrolyse urea. They cause non-gonococcal urethritis. It can be resistant to commonly usedantibiotics such as tetracycline and doxycycline. Other options: Option B - Orientia tsutsugamushi belongs to the family rickettsiaceae. It causes scrub typhus,which is spread by mites. Option C - Chlamydophila pneumoniae belongs to the genus Chlamydia. It is a common cause ofrespiratory diseases such as sinusitis, bronchitis, and pneumonia. Option D - Chlamydia trachomatis also belongs to the genus Chlamydia. It causes trachoma,inclusion conjunctivitis, and genital chlamydiasis.

  5. Question 44 · Microbiology

    A microorganism is resistant to meropenem and aminoglycoside and sensitive toticarcillin-clavulanate, and cotrimoxazole. What is the likely microorganism?

    1. Stenotrophomonas maltophila
    2. Burkholderia cepacia
    3. Pseudomonas aeruginosa
    4. Acinetobacter baumannii
    Show answer

    Answer: A. Stenotrophomonas maltophila

    Stenotrophomonas maltophila, formerly called Pseudomonas maltophila, is an opportunisticpathogen that causes wound infection, urinary tract infection, and septicemia. It is sensitive tocotrimoxazole. Intrinsic resistance is the innate ability of a bacterial species to resist the activity of a particularantimicrobial agent through its inherent structural or functional characteristics, which allowtolerance of a particular drug or antimicrobial class. Burkholderia cepacia, formerly called Pseudomonas cepacia, is an opportunistic pathogen thatcauses fatal necrotizing pneumonia. B. cepacia complex strains are intrinsically resistant toaminoglycosides, polymyxin, and first and second-generation cephalosporins but sensitive tomeropenem, co-trimoxazole, doxycycline, minocycline, chloramphenicol and ceftazidime.

  6. Question 45 · Microbiology

    In which of the following organisms' genetic material is adenine not equal to thymine?

    1. HIV
    2. E coli
    3. Human mitochondrial cells
    4. Herpes virus
    Show answer

    Answer: A. HIV

    The genetic material of HIV consists of a single-stranded RNA and therefore adenine will not beequal to thymine. The pyrimidine components of RNA can differ from those of DNA. Although RNA contains theribonucleotides adenine (A), guanine (G), and cytosine (C), it contains uracil (U) instead ofthymine (T). Because RNA is a single strand that is complementary to only one of two strands of a gene, itsadenine content will not be equal to its uracil content and its guanine will not be equal to itscytosine content. Chargaff's rule states that double-stranded DNA, such as that found in E. coli, humanmitochondrial cells and herpes virus, will always have equal concentrations of (A = T) and (G = C). The pairings of purine and pyrimidine nucleotides on opposite strands are very specific and relyon hydrogen bonds between A and T and G and C. The base pairs A-T and G-C are frequentlyreferred to as Watson-Crick base pairs.

  7. Question 46 · Microbiology

    Which of the following pathological features is incorrectly matched with the disease?

    1. No reaction: Immunodeficiency disease
    2. Cytopathic and cytoproliferative effect : Chickenpox
    3. Mononuclear and granulomatous lesions: Berylliosis
    4. Suppurative and purulent lesions: Syphilis
    Show answer

    Answer: D. Suppurative and purulent lesions: Syphilis

    Syphilis is not known to cause suppurative and purulent lesions. It causes macular, papular ormixed maculopapular lesions. Syphilis presents with the following lesions as it progresses through different stages : • Primary syphilis - Hard chancre, which is painless papule that becomes eroded and indurated. • Secondary Syphilis - Skin rashes on palms and soles which may be macular, papular,papulosquamous and nodular, but never vesicular or bullous. • Condylomata lata- mucocutaneous papules that coalesce to form large pink to grey lesions inwarm moist intertriginous areas) • Generalized lymphadenopathy is seen • Tertiary Syphilis - Gummas, which are locally destructive granulomatous lesions. They are mostcommonly seen in bone and skin. Other options : Option A - Immunodeficient individuals lack the normal response to antigens. This is the reasonfor the false negative Mantoux test seen in such individuals. Option B - Chickenpox belongs to the alpha subfamily of herpes group of viruses which has acytopathic effect of formation of multinucleated giant cells, with rounding and ballooning ofinfected cells Option C - Berryliosis is an occupational lung disease, causing a chronic granulomatousinflammatory disease similar to sarcoidosis. Biopsy reveals non-caseating granulomas withmonocytic infiltration.

Obstetrics and Gynecology

7 questions
  1. Question 47 · Obstetrics and Gynecology

    What should be the minimum percentage for normal sperm morphology in a seminal fluidanalysis?

    1. 10%
    2. 4%
    3. 6%
    4. 8%
    Show answer

    Answer: B. 4%

    4% is the minimum percentage for normal sperm morphology in a seminal fluid analysis. Morphology of sperms is the most important marker of male fertility. Characterization of normalsperm morphology shows improved correlation with fertilization rates during IVF cycles. The criteria requires careful analysis of the following: • The shape and size of the sperm head, • The relative size of the acrosome in proportion to the head, • And characteristics of the tail, including length, coiling, or presence of two tails. Significantly decreased fertilization rates are observed when normal morphology of the samplefalls below 4 percent. Abnormal sperm morphology is termed teratospermia or teratozoospermia.

  2. Question 48 · Obstetrics and Gynecology

    An 18-year-old girl presented with amenorrhoea. On examination, she had normal genitalia& normal breast development. Her tests reveal a 5 alpha-reductase deficiency. Whatcould be the probable diagnosis?

    1. Concealed sex
    2. True hermaphroditism
    3. Female pseudohermaphroditism
    4. Male pseudohermaphroditism
    Show answer

    Answer: D. Male pseudohermaphroditism

    A girl presenting with amenorrhea, normal female external genitalia and breasts with 5alpha-reductase deficiency is suggestive of male pseudohermaphroditism. In pseudohermaphroditism, the sex glands are of a certain sex while the external genitalia is ofthe opposite sex. Male pseudohermaphroditism or male intersex is an association of male gonads (XY genotype)with female external genitalia. The production of testosterone from the testes requires 5alpha-reductase. This enzyme converts testosterone into DHT (dihydrotestosterone), which iscapable of acting on peripheral target tissues to produce the male phenotype. The absence of thisenzyme results in failure of the testis to descend to the scrotum and ill development of the penis,thereby resulting in the female phenotype. Many individuals develop their attitudes and psycheaccording to their sex of rearing (as a female). Rehabilitation to their gender roles is the best lineof management. Other options: Option A: Concealed sex refers to the deliberate concealment of sex usually done by criminals bychanging clothes or by other methods to avoid detection. This can be detected by physicalexamination. Option B: In true hermaphroditism the glands of both the sexes are present in the sameindividual. Such cases are very rare. In most cases, the accessory sex gland is atrophic and lacksfunctional activity. Option C: Female pseudohermaphrodites are genetic females (XX genotype) with male/ambiguousexternal genitalia due to the deficiency of 21 alpha-hydroxylase enzymes resulting in an excess ofandrogens. This causes clitoromegaly resembling the penis,and the fused labioscrotal foldsresemble a rudimentary scrotum. Management includes administration of steroids to controlexcess ACTH, estrogen to confer the secondary sexual characteristics, and surgical restoration ofthe external genitalia to a feminine pattern.

  3. Question 49 · Obstetrics and Gynecology

    A 25-year-old lady had an uneventful vaginal delivery. On postpartum day 1, she presentedwith postpartum haemorrhage and complains of visual changes and severe headache. A fewhours after the complaint, she is found to be in an unresponsive state and is intubated andput on ventilation. On further examination, she has a BP of 60/40 mmHg and tachycardia,blood sugar is 48 mg/dL. Hematocrit and WBC count was normal. Which of the followingdrugs would help reverse this state?

    1. Inj. dobutamine
    2. Steroids
    3. Thyroxine
    4. Activated protien C
    Show answer

    Answer: B. Steroids

    The given history of a pregnant having postpartum hemorrhage developing headache, visualchanges, and hypotension after a vaginal delivery suggests Sheehan syndrome. Steroids are thedrug that can reverse this state. Volume depletion and low cardiac output leading to critically low BP are common in acutepituitary deficiency and recovery of normal cardiovascular status is rapidly achieved underhormone and volume replacement therapy without the need for inotropes like dobutamine. Pituitary apoplexy is a hemorrhagic infarction of large pituitary adenomas or pituitaryinfarction due to traumatic major blood loss. If this blood loss occurs during delivery, it'scalled Sheehan's syndrome. The hyperplastic enlargement of the pituitary (size of gland increasesby about 125%), which occurs normally during pregnancy, increases the risk of hemorrhage andinfarction. Pituitary apoplexy presents acutely as an endocrine emergency, with sudden onset of headache,nausea, visual deficits, persistent hypotension, tachycardia, and hypoglycemia leading to shock.The most prominent symptoms are lactation failure (due to a deficiency ofprolactin) and postpartum amenorrhea (due to a deficiency of gonadotropins). Because deficiencies of some or all pituitary-responsive hormones may develop after the initialinsult, Sheehan syndrome can be heterogeneous and not be identified for years. It can present latewith the following features like loss of pubic and axillary hair, hypothyroidism due to decreasedTSH secretion, adrenal insufficiency due to low ACTH. Treatment is by hormone replacement - first glucocorticoids, later - thyroid hormone, sex steroids,vasopressin, and growth hormones are given. If there is evidence of progressive visual loss, cranialnerve involvement, or loss of consciousness, immediate surgical decompression must be carriedout.

  4. Question 50 · Obstetrics and Gynecology

    Which of the following is false regarding management of diabetes in pregnancy?

    1. Elective C-section has no role in reducing incidence of brachial plexus injury
    2. In a patient being planned for induction of labor, night dose of intermediate insulin is given asplanned, and the morning dose is withheld
    3. In active labor, if RBS <70 mg%, D5 is started at 100-150 ml/hr till the RBS is >100 mg%
    4. Capillary blood glucose monitoring levels are kept at fasting- 95 mg%; 1 hr postprandial- 140mg%; 2 hrs postprandial- 120 mg%
    Show answer

    Answer: A. Elective C-section has no role in reducing incidence of brachial plexus injury

    False statement regarding management of diabetes in pregnancy is option A. Cesarean section isfound to be protective against traumatic injuries such as brachial plexus injury, in macrosomicbabies born to mothers with diabetes. An elective C-section is found to be more protective than anemergency C-section. Almost 50 % of pregnant women with diabetes are delivered by c-section. Indications forc-section in diabetic pregnant women are- • Fetal macrosomia of ≥4.5kgs to avoid trauma to mother or neonate, and also to avoid brachialplexus injury in the baby • Unsatisfactory progression of labor • Complicated diabetes mellitus • Fetal compromise during delivery • Elderly primigravida (age >30 years) • Multigravida with bad obstetric history previously • Pregnancy that is associated with complications like preeclampsia, polyhydramnios ormalpresentation Other options: Option B: In a planned delivery, although the bedtime dose of intermediate-acting insulin is given,it is recommended that the morning dose must be withheld. This is due to reduced insulinrequirements after delivery. Therefore, regular insulin is used for glycemic control during thistime. Option C: In active labor, if RBS (Random blood glucose) <70 mg%, D5 (5% dextrose) is startedat 100-150 ml/hr till the RBS is >100 mg%. Before initiation of labor, normal saline is used foriv infusion. Option D: Self-monitoring of capillary blood glucose is advised in a pregnant woman diagnosedwith diabetes. Subcutaneous continuous glucose monitoring devices can also be used for thesame. The glucose goals recommended are: • Fasting: ≤95 mg/dL • 1 hour post-prandial: ≤140 mg/dL • 2 hrs post-prandial: ≤120 mg/dL

  5. Question 51 · Obstetrics and Gynecology

    SRY region is located in:

    1. Short arm of Y chromosome
    2. Short arm of X chromosome
    3. Long arm of Y chromosome
    4. Long arm of X chromosome
    Show answer

    Answer: A. Short arm of Y chromosome

    SRY region is present on the short arm of the Y chromosome(Yp11). SRY stands for the sex-determining region on Y. It contains the SRY gene, which produces theprotein called the testis determining factor. This causes the gonads to develop into the testes. Ifthe SRY gene or Y chromosome is absent, the gonads develop as ovaries.

  6. Question 52 · Obstetrics and Gynecology

    In a 46XY female complete gonadal dysgenesis was noted. Complete gonadal dysgenesis iscaused by what in the SRY gene?

    1. Point mutation
    2. Deletion of gene
    3. Translocation
    4. Inversion
    Show answer

    Answer: A. Point mutation

    Amongst the given options, a point mutation in the SRY gene is most likely to result in completegonadal dysgenesis but this answer remains controversial as some other sources mention SRYdeletion is more commonly associated. The most common cause attributable to about 85% ofcases is still idiopathic. Pure gonadal dysgenesis results from a point mutation in SRY/ SRY deletion or point mutationsin another gene with testis-determining effects (DAXJ, SF-1, CBX2). This leads to underdevelopeddysgenetic gonads that fail to produce androgens or AMH. This is characterized by a normalprepubertal female phenotype and a normal mullerian system due to absent AMH. The child usually presents with failure in the development of secondary sexual characteristics atpuberty and amenorrhea. Investigations reveal the absence of estrogen production and they haveelevated levels of FSH (>40 miU/mL) and low oestradiol levels (<25 pg/mL). Women with streak ovaries are infertile, but they can bear children with oocyte donation. Theymust be treated with cyclic estrogen and progestogen to promote feminization and secondarysexual characteristics and prevent osteoporosis.

  7. Question 53 · Obstetrics and Gynecology

    An 18-year-old girl was brought by her mother with complaints of primary amenorrhea andacute retention of urine. She had cyclical abdominal pain and on examination, a supra pubic mass and a tense bluish bulge were seen in her vagina. What is the most appropriate cause ofher symptoms?

    1. Imperforate hymen
    2. Transverse vaginal septum
    3. Vertical vaginal septum
    4. Mullerian agenesis
    Show answer

    Answer: A. Imperforate hymen

    The features of primary amenorrhea, cyclical abdominal pain, acute retention of urine along witha tense bluish bulge make the diagnosis of imperforate hymen more likely. Imperforate hymen is a type of Mullerian defect, of urogenital origin. Failure of canalization leadsto the formation of mucocolpos; this may be recognized in early infancy. However, the anomalyoften continues unrecognized until puberty. It presents with amenorrhoea in the presence ofsecondary sexual characters, cyclic abdominal discomfort, retention of urine and a suprapubicbulge caused by hematocolpos. Examination of the external genitalia reveals the presence of atelltale bluish bulging membrane in the region of the hymen. A simple cruciate incision followed by an excision of the hymenal tags allows the drainage of theretained menstrual blood. The operation should be performed under aseptic conditions and underan adequate antibiotic cover to avoid any ascending infection. Other options : Option B - Transverse vaginal septum is a much rarer condition that occurs due to faulty fusion orcanalization of the urogenital sinus and the Müllerian ducts. The majority of the cases involve theupper portion of the vagina. It can be differentiated from imperforate hymen by the presence of ashort vagina and the absence of a bulging membrane. Option C - Vertical vaginal septum occurs when the distal parts of the Müllerian ducts fail to fuse.On examination, there is a longitudinal anteroposterior septum which may be partial or complete,extending right down to the vaginal outlet. It may be associated with a double uterus and doublecervix. Option D - Mullerian agenesis is characterized by congenital absence of the uterus and vagina(small rudimentary uterine bulbs are usually present). The female presents with primaryamenorrhea and preservation of female secondary sexual characteristics.

Ophthalmology

6 questions
  1. Question 54 · Ophthalmology

    The most common tumor of the lacrimal gland is?

    1. Adenoid cystic carcinoma
    2. Non Hodgkin Lymphoma
    3. Pleomorphic adenoma
    4. Mucoepidermoid carcinoma
    Show answer

    Answer: C. Pleomorphic adenoma

    Pleomorphic adenoma is the most common tumor of the lacrimal gland while adenoid cysticcarcinoma is the most common carcinoma of the lacrimal gland.

  2. Question 55 · Ophthalmology

    SAFE strategy for trachoma includes all except?

    1. Surgery for trichiasis
    2. Antibiotics
    3. Facial cleanliness
    4. Evaluation of control program
    Show answer

    Answer: D. Evaluation of control program

    E in the SAFE strategy stands for Environmental improvement with better hygiene and sanitation.

  3. Question 56 · Ophthalmology

    What is the most common cause of moderate-severe visual impairment according to theNational Blindness and Visual Impairment Survey 2015-19?

    1. Refractive error
    2. Untreated cataract
    3. Glaucoma
    4. Diabetic retinopathy
    Show answer

    Answer: B. Untreated cataract

    According to the National Blindness and Visual Impairment Survey 2015-19, the most commoncause of moderate-severe visual impairment (MSVI) is untreated cataract. MSVI is defined as presenting visual acuity of <6/18 to 3/60 in the better eye with availablecorrection. Most commonly, cases of visual impairment and blindness are more prevalent in patients aged 50years and above. Other options: Option A: Refractive errors are the leading cause of MSVI in patients aged 49 years and below. Options C and D: Glaucoma and diabetic retinopathy account for a very low number of visualimpairment cases in patients aged 50 years and above.

  4. Question 57 · Ophthalmology

    A patient has presented for a routine eye evaluation. You have checked his visual acuity usingSnellen's chart and found it to be 6/6. What is the minimum angle of resolution?

    1. 15 minutes of arc
    2. 5 minutes of arc
    3. 10 minutes of arc
    4. 20 minute of arc
    Show answer

    Answer: B. 5 minutes of arc

    The angle subtended for a patient with 6/6 vision is 5 minutes of arc. Visual acuity is a measure of the spatial resolution of the eye or an estimation of its ability todiscriminate between two points. Snellen's chart consists of a series of letters arranged in lines, each diminishing in size.Each whole letter will subtend an the angle of 5 minutes at the nodal point of the eye at the given distance. In the Snellen chart, the largest letter will subtend an angle of 5 minutes at the nodal point if it is60 m from the eye. Those in the subsequent lines will subtend an angle of 5 minutes if they are36, 24, 18, 12, 9, and 6 m from the eye. The patient is placed at 6 m and asked to read the visual acuity chart from the largest lettersdownwards to the smallest that can be visualized with each eye separately. A person with normal6/6 vision sitting at a distance of 6 m from the types should be able to read every letter from thetop to the end of the 6 m line. So the angle subtended by the 6 m line at the eye a distance of 6meters will be 5 minutes of arc. If a patient can only read up to the 12m line, his distant vision is defective. His visual acuity willbe recorded as 6/12. The numerator denotes the distance at which the patient is able to read theparticular line and the denominator denotes the distance at which a person with normal visualacuity will be able to read the given line.

  5. Question 58 · Ophthalmology

    A construction site worker comes in with red eyes and lacrimation. He got chuna particles inhis eyes an hour ago. What is the initial step in the treatment?

    1. Irrigation with normal saline
    2. Topical antibiotics and cycloplegics
    3. Patch the eye after topical lubricants
    4. Temporary tarsoraphy
    Show answer

    Answer: A. Irrigation with normal saline

    The initial step in the management of a patient presenting with chemical burns is immediate andthorough irrigation with the available clean water or normal saline delivered through an IVtubing. This is done to prevent further damage. Along with irrigation, contaminant elimination through mechanical means shall be done. In theevent of residue, especially with lime (chuna), it is essential to meticulously eliminate them usinga swab stick. After the initial steps, maintenance of favorable conditions for rapid and uncomplicated healingby following measures: • Topical antibiotic drops, for example, moxifloxacin 4-6 times a day to prevent infection. • Steroid eye drops to decrease inflammation, neutrophil infiltration, and treat anterior uveitis.Caution is advised after 2 weeks, as steroids could disrupt collagen synthesis. • Cycloplegics, such as atropine, might enhance comfort. Grading of severity (Roper–Hall system) of chemical injuries is based on corneal clarity andseverity of limbal ischemia: • Grade 1 is characterized by a clear cornea (epithelial damage only) and no limbal ischemia(excellent prognosis). • Grade 2 shows a hazy cornea but with visible iris detail and less than one-third of the limbusbeing ischemic (good prognosis). • Grade 3 manifests as a total loss of the corneal epithelium, stromal haze obscuring iris detail,and between one-third and half of the limbus showing ischemia (guarded prognosis). • Grade 4 manifests with an opaque cornea and more than 50% of the limbus showing ischemia(poor prognosis).

  6. Question 59 · Ophthalmology

    The following are features of third cranial nerve palsy except

    1. Elevation
    2. Adduction
    3. Limited abduction
    4. Ptosis
    Show answer

    Answer: C. Limited abduction

    Limited abduction is not a feature of third cranial nerve palsy. Abduction is carried out by the lateral rectus, which is supplied by the abducens (sixth) nerve. Third cranial nerve (oculomotor nerve) palsy presents with: • Profound ptosis due to the weakness of the levator muscle • Limited adduction due to medial rectus weakness • Limited elevation due to the weakness of the superior rectus and inferior oblique • Abduction and depression in the primary position ("down and out")

Orthopedics

3 questions
  1. Question 60 · Orthopedics

    Which of the following findings appear late in compartment syndrome?

    1. Pallor
    2. Paralysis
    3. Pulselessness
    4. Pain on passive strech
    Show answer

    Answer: C. Pulselessness

    Pulselessness is a late sign is compartment syndrome. Acute compartment syndrome is a surgical emergency caused by increased pressure within afascial compartment, leading to tissue ischemia. It is characterized by rapid progression of painand swelling in an extremity, commonly precipitated by traumatic injury. The classic features of ischemia in acute compartment syndrome include the 5 P's: • Pain: Pain is the first symptom of acute compartment syndrome. Ischemic muscle groups havean increased sensitivity to touch. On passive stretching of the muscles, severe pain can be elicited. • Paresthesia: Paresthesia and hypoesthesia are the first signs of ischemia and may be noted dueto compression of nerves traversing the affected compartment. • Pallor: Pallor of the skin may not always be seen as ischemia occurs at the capillary level (latesign). • Paralysis: Paralysis of muscle groups affected in that compartment (late sign). • Pulselessness: The absence of distal pulses (extremely late sign). Treatment: • Prompt decompression of the affected compartment is done. • All the dressings, bandages, and cast on the affected limb are removed and the limb is nursedflat. • Fasciotomy is done if the differential pressure is less than 30 mm Hg • Fasciotomy wounds are left open and after 2 days if there is necrosis debridement is done. If thewounds are healthy it can be sutured without tension or skin grafting can be done. The diagnosis of compartment syndrome can be confirmed by measuring the intra-compartmentalpressure using a split catheter introduced into the compartment. The difference between diastolicpressure and compartment pressure is known as differential pressure (∆P). If the differentialpressure (∆P) is less than 30 mm Hg (4.00 kilopascals), immediate compartmentdecompression is indicated.

  2. Question 61 · Orthopedics

    A 65-year-old postmenopausal female patient comes with persistent backache not respondingto conservative treatment. She has a history of lifting heavy weights 4 months back andprogressively increasing pain that worsens on walking. The pain starts radiating down herlegs after walking for a distance of around 100 m. However, she feels no pain on walkinguphill. What is the most probable diagnosis?

    1. Osteoporotic compression fracture
    2. Lumbar stenosis
    3. Thromboangiitis obliterans
    4. Atherosclerosis
    Show answer

    Answer: B. Lumbar stenosis

    The given clinical scenario of an old woman with back pain that worsens on walking and relieveswalking uphill points to the diagnosis of lumbar stenosis. Walking uphill may be preferred as it flexes the spine and maximizes the spinal canalcapacity, relieving the pain. Spinal stenosis is the narrowing of the spinal canal, causing nerve compression. It can becongenital or acquired, with degeneration being the most common cause. Other factors includeconditions like spondylolysis, trauma, infection, and metabolic disorders. Degenerative spinal stenosis involves disc degeneration, facet joint hypertrophy, and ligamentumflavum calcification. These changes lead to reduced spinal canal dimensions and compression ofthe nerves resulting in neurogenic claudication. They present with symptoms such as back pain (most common), sciatica, sensory disturbances inthe legs, motor weakness, and sometimes urinary disturbances. Pain typically occurs afterstanding or walking for a few minutes and is relieved by sitting, squatting, or flexing the spine. MRI is the imaging method of choice for the investigation of spinal stenosis. Conservativetreatment is the first line for spinal stenosis, including reassurance, pain relief, exercise, andactivity modification. Surgery is considered if conservative measures fail and there is significant functional impairmentand severe neurological symptoms. Posterior decompression is the surgical method. Laminectomyis performed for central stenosis.

  3. Question 62 · Orthopedics

    Which of the following layers is affected in Salter Harris type 1 fracture?

    1. Through the whole of epiphysis
    2. Resting zone of the growth plate
    3. Calcific layer of the growth plate
    4. Mineralisation zone
    Show answer

    Answer: C. Calcific layer of the growth plate

    Salter Harris type 1 fracture is a transverse fracture through the hypertrophic or calciIed zone ofthe growth plate. The growing zone of the physis is usually not injured and growth disturbance isuncommon.

Paediatrics

6 questions
  1. Question 63 · Paediatrics

    Which can be used as an alternative to mother's milk in a low birth weight (LBW) baby?

    1. Cow's milk
    2. Preterm formula
    3. Term formula
    4. Human donor milk
    Show answer

    Answer: D. Human donor milk

    Human donor milk is the best alternative to mother's milk for low birth weight (LBW) babies. It is recommended that all LBW babies are exclusively fed breast milk regardless of the feedingmethod. This can be achieved through the use of expressed breast milk (EBM) from the mother orhuman donor milk. Donor milk offers advantages over preterm formula, such as reduced risk of necrotizingenterocolitis (NEC) and feeding intolerance. However, due to lower protein and micronutrientlevels in donor milk, human milk fortifiers should be added to meet the nutritional needs ofinfants. In cases where breastfeeding is not feasible or the mother is ill, formula feeds can be utilized.Term formula feed (option C) is suitable for infants weighing over 1500 grams at birth, whilepreterm formula feed (option B) is recommended for very low birth weight infants. Cow milk (option A) is not preferred as it lacks essential nutrients such as vitamin C, iron, andessential fatty acids. Additionally, its high protein and lactose content can lead to digestive issuesand allergic reactions in infants. Once the mother's condition stabilizes or breastfeeding contraindications are resolved, exclusivebreastfeeding should be resumed. The mode of nutrition of LBW depends on their gestational age. Babies born before 28 weeks ofgestation receive total parenteral nutrition (TPN). Enteral feeding is not initiated at this stage asthe gastrointestinal tract is not fully developed. Babies born between 28 and 32 weeks of gestation may exhibit suckling bursts but theircoordination of suckling, swallowing, and breathing is not fully developed. Hence, EBM givenorally may result in aspiration. EBM is administered through a nasogastric tube or orogastrictube. As gestational age advances to 32-34 weeks, the coordination of feeding improves, and EBM canbe given with a paladai or Katori spoon. For babies born after 34 weeks of gestation, with fully developed coordination, directbreastfeeding is possible.

  2. Question 64 · Paediatrics

    Handedness is seen at the age of ______.

    1. 1 year
    2. 2 years
    3. 3 years
    4. 4 years
    Show answer

    Answer: C. 3 years

    Handedness is seen at the age of 3 years. The child may become frustrated if attempts are made inchanging the hand preference.

  3. Question 65 · Paediatrics

    Which of the following has the least risk of perinatal transmission?

    1. HSV
    2. CMV
    3. Rubella
    4. Hepatitis B
    Show answer

    Answer: C. Rubella

    Rubella has the least risk of perinatal transmission. Rubella or German measles has an intrauterine transmission that causes serious infection inpregnancy which results in abortions or severe fetal defects if infection occurs in the firsttrimester. The classical triad of congenital rubella syndrome is deafness, cataract, and congenitalheart disease. Other options- Option A: Herpes simplex infection can be acquired in utero, intrapartum/perinatally, or duringthe neonatal period. The intrauterine infection causes skin vesicles, eye findings (chorioretinitis,keratoconjunctivitis), and microcephaly/hydranencephaly. Intrapartum infection presents witheither: • Skin, eye, and mouth disease • Encephalitis • Disseminated disease with sepsis-like presentation Option B: Cytomegalovirus, the most common congenital infection with perinatal transmissionpresents with hepatosplenomegaly, petechial rashes, jaundice, chorioretinitis, and microcephaly.One of the long-term sequelae is hearing loss. Option D: Hepatitis B virus is associated with significant perinatal transmission. The mostimportant risk factor that determines perinatal transmission is HBeAg.

  4. Question 66 · Paediatrics

    A 3.5 kg term neonate born to a mother with eclampsia, did not cry at birth. Which of thefollowing is the criteria for perinatal asphyxia?

    1. APGAR persists at 4-7 for > 5 minutes
    2. Cord pH < 7.2
    3. Hypocalcemia
    4. Hypotonia
    Show answer

    Answer: D. Hypotonia

    Among the given options, the essential criterion for perinatal asphyxia is hypotonia. Perinatal asphyxia is an insult to the fetus or newborn due to a lack of oxygen (hypoxia) and/or alack of perfusion (ischemia) to the various organs. It is often associated with tissue lactic acidosisand hypercarbia. Essential criteria for perinatal asphyxia: • Umbilical arterial blood sample indicating prolonged metabolic or mixed acidemia (pH < 7.O). • Apgar score of 0-3 persisting for >5 minutes. • Neurological manifestations appearing in the immediate neonatal period Ex:- seizures, coma,hypotonia, or hypoxic ischemic encephalopathy (HIE). • Neonatal multiorgan dysfunction.

  5. Question 67 · Paediatrics

    What is the ideal maintenance fluid of choice in children?

    1. 5% dextrose and N/2 saline
    2. 5% dextrose and N saline
    3. 10% dextrose and N/2 saline
    4. 5% dextrose and N/4 saline
    Show answer

    Answer: B. 5% dextrose and N saline

    5% dextrose and N saline is the ideal maintenance fluid in children. 5% dextrose provides 17 calories/100 mL and nearly 20% of the daily caloric needs. This level isenough to prevent ketone production and helps minimise protein degradation. However, the childwill lose weight on this regimen and therefore it is necessary to start enteral feeds sooner or if notpossible, total parenteral nutrition is recommended. Other options: Hypotonic fluids like N/2 or N/4 saline can increase the risk of hyponatraemia and free watermovement inside RBCs, leading to haemolysis. Hence, isotonic fluids with 5% dextrose arerecommended as standard maintenance fluids in children.

  6. Question 68 · Paediatrics

    All are true about renal tubular acidosis in children except

    1. Type 4 is due to defective ammoniagenesis and aldosterone resistance
    2. Type 2 is Fanconi and hypophosphatemic rickets
    3. Type 4 with paradoxical aciduria and hypokalemia
    4. Type 2 - abnormal loss of HCO3
    Show answer

    Answer: C. Type 4 with paradoxical aciduria and hypokalemia

    Hyperkalemia and metabolic acidosis are features of type 4 renal tubular acidosis Type IV (hyperkalemic) renal tubular acidosis: Hyperkalemia with distal RTA occurring due toaldosterone resistance or deficiency is termed type 4 RT A. Aldosterone directly stimulates theproton pump, increases Na+ absorption resulting in negative intrahibular potential and increasesurinary K+ losses and stimulates basolateral Na+ /K+ ATPase. Hence, aldosterone deficiency orresistance is expected to cause hyperkalemia and acidosis. The clinical features seen in Type IV RTA are as follows: • Hyperkalemia: Hypoaldosteronism results in an increase in potassium concentration. • Metabolic acidosis: The important buffers in urine are ammonia and phosphate. Hyperkalemiaimpairs ammonia production in the proximal tubule, which reduces the buffering of hydrogenions. This results in decreased hydrogen ion excretion. • Urinary pH<5.5: Due to the decreased production of ammonia in the proximal tubule, excessprotons are excreted in the renal tubules, thus decreasing the urinary pH. RTA type I (distal RTA) is seen in autoimmune diseases like Sjogren's syndrome, systemic lupuserythematosus, systemic sclerosis, medullary sponge kidney disease, cirrhosis, drugs likeamphotericin B, lithium, etc. The patients develop hypokalemia and metabolic acidosis withalkaline urine (urinary pH >5.5). RTA type II (proximal RTA) is seen in Fanconi syndrome, Wilson's disease, heavy metalsingestion, etc., The patients develop hypokalemia and metabolic acidosis with acidic urine(urinary pH <5.5). It occurs due to generalized proximal tubule dysfunction. Defective HCO3-resorption is the primary dysfunctional mechanism causing metabolic acidosis and is alsoassociated with glycosuria, aminoaciduria, and phosphaturia. Rickets is rare in isolated proximalRTA but common in Fanconi syndrome. RTA type III (mixed RTA) is seen in inherited mutations of carbonic anhydrase II. A combinedpathology of type I and type II RTA is seen in this type. It results in hypokalemia and metabolicacidosis.

Pathology

7 questions
  1. Question 69 · Pathology

    What is the cell of origin of Chronic Lymphocytic Leukaemia/Small Lymphocytic Lymphoma?

    1. Mature B cells
    2. Mature T cells
    3. Progenitor T cells
    4. Naive B cells
    Show answer

    Answer: D. Naive B cells

    The cell of origin of Chronic Lymphocytic Leukaemia (CLL)/ Small Lymphocytic Lymphoma (SLL)is either naive B cells or post-germinal centre memory B cell. The CLL and SLL differ only in the degree of peripheral blood lymphocytes. CLL is the mostcommon leukemia of adults in the western world. The median age of diagnosis is 60 years. Themost common genetic anomalies seen are deletions of 13q14.3, 11q, and 17p and trisomy 12q. Lossof tumor suppressor genes in chromosome 13 results in overexpression of the anti-apoptoticprotein BCL2. The growth of CLL/SLL cells is confined to the proliferation centers (pathognomonic forCLL/SLL), where the tumor cells receive critical factors from the surrounding environments, suchas NF-κB for cell survival, MYC for cell growth, Bruton Tyrosine Kinase (BTK) signaling pathway,etc., The tumor cells express CD19, CD20, CD23, and CD5. Smudge cells are also characteristic ofCLL. Clinical features include easy fatiguability, weight loss, anorexia, generalizedlymphadenopathy, hepatomegaly, hemolytic anemia, etc. CLL/SLL has a tendency to transform toa more aggressive diffuse large B-cell lymphoma (DLBCL) - this is called Richter syndrome. Factors with a bad prognosis include: • Presence of deletions of 11q and 17p. • Lack of somatic hypermutation. • Expression of ZAP-70, a protein that augments signals produced by the Ig receptor. • Presence of NOTCH1 mutations BTK inhibitors and BCL2 inhibitors produce sustained therapeutic responses in a higher fractionsof people with CLL/SLL.

  2. Question 70 · Pathology

    Which of the following can be prevented by transfusing irradiated RBCs?

    1. Graft versus host disease
    2. HLA Alloimmunization
    3. Immunomodulation
    4. Transfusion Related Acute Lung Injury (TRALI)
    Show answer

    Answer: A. Graft versus host disease

    Transfusion of irradiated RBCs prevent the incidence of graft versus host disease (GVHD)associated with the transfusion. GVHD occurs when immunologically competent cells or their precursors are transplanted intoimmunologically compromised recipients and the transferred cells recognize the alloantigens inthe host and attack host tissues. It is seen most commonly during hematopoietic stem celltransplantation but may occur following transplantation of solid organs too. GVHD maybe of two types: acute and chronic. Acute GVHD occurs within days to weeks after allogeneic hematopoietic stem cell transplantation.It mainly involves the immune system and the epithelia of the skin (in the form of generalizedrash), liver( in the form of jaundice), and intestines (in the form of bloody diarrhea). But theaffected tissues are not heavily infiltrated by lymphocytes. Chronic GVHD may follow acute GVHD or may occur insidiously. These patients have extensiveskin damage, with destruction of skin appendages and fibrosis of the dermis, resembling systemicsclerosis. Cholestatic jaundice and esophageal strictures are also common. There will beinvolution of the thymus and depletion of lymphocytes in the lymph nodes. There will also berecurrent bouts of life-threatening infections. Some may even develop autoimmunity, owing to thegrafted CD4+ helper T cells reacting with host B cells and stimulating them, which may result inautoantibodies production. Depletion of donor T cells by irradiation, before transfusion can eliminate the occurrence ofGVHD. But irradiation also has its side effects such as recurrence of tumor in leukemic patients,increased incidence of graft failures and EBV-related B-cell lymphoma.

  3. Question 71 · Pathology

    Which of the following disorders is inherited in an X-linked dominant pattern?

    1. Rett syndrome
    2. Duchenne muscular dystrophy
    3. Huntington disease
    4. Tuberous sclerosis
    Show answer

    Answer: A. Rett syndrome

    Rett syndrome is inherited in an X-linked dominant pattern. Rett syndrome is a neurodevelopmental disorder. It is caused due to mutation in the MECP-2gene. The disorder is mostly seen in girls because affected boys usually die in infancy due to adefective single X chromosome. Symptoms of Rett syndrome are microcephaly, developmental delay, autistic-like behavior,cognitive dysfunction, seizures, and repetitive motor actions. Other options: Option B: Duchenne muscular dystrophy (DMD) is inherited in X linked recessive pattern. Itoccurs due to a mutation in the dystrophin gene. Typical features include calf musclepseudohypertrophy and Gower's sign (using arms to lift themselves from a seated position).Laboratory investigations show elevated serum creatine kinase levels. Casimersen has beenrecently approved by the FDA for the treatment of DMD. Option C: Huntington's disease is inherited in an autosomal dominant pattern. It is a trinucleotide(CAG) repeat disorder and shows genetic anticipation (earlier onset of disease or increasedseverity in succeeding generations). Patients present with involuntary movements known aschorea. Tetrabenazine is used for treatment. Option D: Tuberous sclerosis also known as Bournville disease is inherited in an autosomaldominant pattern. It occurs due to mutation in TSC1 and TSC2 genes. It is characterized bysubependymal nodules, angiomyolipomas, cafe au lait spots, shagreen patches, and cardiacrhabdomyomas.

  4. Question 72 · Pathology

    Which of the following cytokines is not involved in Th1 response?

    1. IFN gamma
    2. IL- 2
    3. IL- 12
    4. IL- 4
    Show answer

    Answer: D. IL- 4

    The cytokine not involved in TH1 response is IL-4. Dendritic cells and macrophages present antigens to T helper cells (TH) after which theydifferentiate into TH1 and TH2 cells. They are responsible for cell-mediated and humoral immuneresponses, respectively. IL-12 is secreted by macrophages and promotes the proliferation of TH1cells. TH1 cells secrete: • IL-2 which activates natural killer cells (NK cells) and cytotoxic T cells • IFN gamma which stimulates macrophages and is also responsible for delayed hypersensitivityalong with TGF beta. TH2 cells secrete: • IL- 4 stimulates IgE production from B cells • IL- 5 stimulates proliferation of eosinophils and together play a role in helminthic and allergicreactions • IL- 6 promotes proliferation of B cells and production of antibodies.

  5. Question 73 · Pathology

    A patient on long term hemodialysis due to renal failure is likely to develop which type ofamyloidosis?

    1. Amyloid light chain (AL)
    2. Amyloid A (AA)
    3. Amyloid β2 microglobulin (Aβ2M)
    4. Amyloid transthyretin (ATTR)
    Show answer

    Answer: C. Amyloid β2 microglobulin (Aβ2M)

    Patients on long-term hemodialysis for renal failure can develop amyloidosis as a result of thedeposition of β2-microglobulin. β2-microglobulin is a component of the MHC class I molecule and a normal serum protein. In thepast, it was retained in the circulation of patients with chronic kidney disease because it could notbe filtered through dialysis membranes. In these patients, it would get deposited in the joints,muscles, tendons, and ligaments. One of the complications which can occur as a result is carpaltunnel syndrome. However, with the advent of new dialysis filters, this complication has becomeuncommon now.

  6. Question 74 · Pathology

    Most common genetic cause of hypercoagulability is ______

    1. Factor V Leiden mutation
    2. Protein C deficiency
    3. Antithrombin III deficiency
    4. G20210A prothrombin gene mutation
    Show answer

    Answer: A. Factor V Leiden mutation

    The most common genetic cause of hypercoagulability is the Factor V Leiden mutation. Primary (inherited) hypercoagulability is most often caused by mutations in the factor V andprothrombin genes. Factor V Leiden mutation is the most common cause followedby prothrombin G20210A mutation. The Leiden mutation is inherited in an autosomal dominant pattern. The mutation makes factor Vresistant to cleavage and inactivation by protein C. This leads to the inactivation of an importantantithrombotic counterregulatory pathway. Protein C, Protein S, and antithrombin III deficiency are rare causes of primaryhypercoagulability. Other options: Option B: Protein C deficiency causes thrombosis in teenagers. It may also be induced by liverdysfunction, vitamin K antagonists, renal failure, DIC, and active thrombosis. Option C: Antithrombin III forms a complex with thrombin by binding to heparin on endothelialcells. This complex inhibits coagulation. Its deficiency presents as early-age thrombosis (less than50 years old) and it carries the highest risk for thrombotic events among the inheritedthrombophilias. Option D: The prothrombin G20210A mutation results in increased levels ofprothrombin which demonstrates a higher risk for arterial and venous thrombotic events.

  7. Question 75 · Pathology

    Aneurysm due to excessive degradation of connective tissue is seen in which of thesedisorders?

    1. Marfan's syndrome
    2. Ehler Danlos syndrome
    3. Scurvy
    4. Atherosclerosis
    Show answer

    Answer: A. Marfan's syndrome

    Aneurysm due to excessive degradation of connective tissue is seen in Marfan's syndrome. Marfan's syndrome arises from a mutation in the FBN1 gene, responsible for encoding fibrillin, acrucial protein essential for maintaining the structural integrity of connective tissues andregulating TGF-β signaling. The condition primarily affects the skeleton, eyes, and cardiovascularsystem. Clinical manifestations commonly include tall stature, elongated fingers, bilateral lenssubluxation, mitral valve prolapse, aortic aneurysm, and potentially life-threatening aorticdissection. Cardiovascular complications pose the greatest threat to this disorder. The two primary concernsare mitral valve prolapse and dilation of the ascending aorta due to cystic medial necrosis.Progressive dilation of the aortic root and valve ring results from the loss of medial support,leading to severe aortic incompetence. Additionally, medial weakening increases the risk ofintimal tears, which can initiate intramural hematomas. These hematomas may cleave throughlayers of the aortic media, potentially causing aortic dissection. In some cases, the hemorrhageextends extensively, from the aortic root to the iliac arteries and may ultimately rupture throughthe aortic wall.

Pharmacology

7 questions
  1. Question 76 · Pharmacology

    A patient who was diagnosed with epilepsy was put on retigabine TDS. Now phenytoin isbeing added. Which of the following changes should be made to retigabine?

    1. Increase the dose
    2. Decrease the dose
    3. Stop the drug
    4. Stop retigabine and start on carbamazepine
    Show answer

    Answer: A. Increase the dose

    The dose of retigabine should be increased when phenytoin is added because phenytoin candecrease the plasma concentration of retigabine when given together. Retigabine is also known as ezogabine. Metabolism of the drug occurs by glucuronidation andacetylation. Its half-life is about 7-11 hours and hence requires thrice-daily dosing. When it isadministered with phenytoin or carbamazepine there may be a reduction in the plasmaconcentration of the drug. So the dose of retigabine should be increased when given along withthese drugs. It is a potassium(K+) channel opener that enhancer transmembrane potassium currents mediatedby the KNCQ family of ion channels. It was approved as an adjunctive agent for focal onsetseizures of patients above 18 years. But because of its adverse effects like retinal abnormalitiesand blue pigmentation, the production of the drug was discontinued by the manufacturer. Other drug interactions of phenytoin: It is an inducer of the enzymes CYP2C8/9, CYP3A4/5, and other CYPs. It is also an inhibitor ofCYP2C9/19. • Phenobarbitone can inhibit the metabolism of phenytoin. But the overall interaction isunpredictable as both the drug enhance each other's degradation by enzyme induction. • Phenytoin and carbamazepine induce each other's metabolism. • Valproate increases the level of unbound phenytoin by displacing the protein-bound form anddecreasing its metabolism. • Phenytoin increases the metabolism of steroids, oral contraceptives, doxycycline, theophylline,etc by enzyme induction. • It produces competitive inhibition of warfarin metabolism. (Increased risk of bleeding whengiven together) • Drugs like chloramphenicol, warfarin, isoniazid, and cimetidine can inhibit the metabolism ofphenytoin and cause its toxicity. • Sucralfate decreases the absorption of phenytoin by binding to it in the gastrointestinal tract.

  2. Question 77 · Pharmacology

    Topiramate is used for

    1. Treatment of migrane
    2. Treatment of Lennox Gastaut syndrome
    3. Prophylaxis of ADHD
    4. Treatment of dementia
    Show answer

    Answer: B. Treatment of Lennox Gastaut syndrome

    Topiramate is used in the treatment of Lennox-Gastuat syndrome. Uses of topiramate: • In newly diagnosed focal and primary generalized epilepsy in adults and children • As monotherapy for refractory focal epilepsy and generalized tonic-clonic seizures • Treatment of tonic-clonic seizures in patients with Lennox-Gastuat syndrome • It is also used in the prophylaxis of migraine(not for treatment) • It can be given as monotherapy for children of 10 years or more. But if used as adjunctivetherapy it can be used in children as young as 2 years. Mechanism of action: It reduces voltage-gated sodium currents in the cerebellar granule cells. Italso activates a hyperpolarizing potassium current, enhances GABAA receptor currents, and limitsthe activation of the AMPA-kainate subtype of glutamate receptors. It also produces weakinhibition of the carbonic anhydrase enzyme. Pharmacokinetics: Only 10-20% is bound to plasma proteins and the drug is largely excretedunchanged in the urine. Half-life is about 1 day. Topiramate is found to decrease the plasmaconcentrations of oral contraceptives, hence a higher dose of OCPs should be given along withtopiramate. Adverse effects: • Somnolence, weight loss, fatigue, and nervousness are commonly seen • It can precipitate renal calculi. It may be due to the weak carbonic anhydrase inhibition. • Cognitive impairment

  3. Question 78 · Pharmacology

    A young boy with a history of multiple blood transfusions presents with bone pain for thepast 6 months. A peripheral smear showed sickle-shaped RBCs. Which of the following drugsare not used in the treatment of his condition?

    1. Bebtelovimab
    2. Voxelotor
    3. Hydroxyurea
    4. L-glutamine
    Show answer

    Answer: A. Bebtelovimab

    Bebtelovimab is not used in the treatment of sickle cell disease. The clinical features of multipleblood transfusions, bone pain, and sickle cells on peripheral smear are suggestive of sickle celldisease. Bebtelovimab is the new monoclonal antibody used for the treatment of COVID-19. It works bybinding to the spike protein of the COVID-19 virus. It has been found useful in reducing the riskof hospitalization or death. Bebtelovimab retains the activity against the omicron variant. Other options: Option B: Voxelotor is an inhibitor of hemoglobin S polymerase and stabilizer of sickle cellhemoglobin (HbS) that is used in the therapy of sickle cell disease. Option C: Hydroxyurea improves fetal haemoglobin (HbF) and is the primary drug for controllingHbS in sickle cell disease. Thus improving the oxygen-carrying capacity of the blood. Hydroxyureaalso inhibits the enzyme ribonucleotide reductase (RNR), which is a rate-limiting step in thebiosynthesis of DNA. This property of hydroxyurea is used in the treatment of various cancers,and myeloproliferative disorders like chronic myelogenous leukemia, and polycythemia. Option D: L-glutamine is a naturally occurring amino acid with antioxidant properties.It increases the availability of reduced glutathione which may lessen oxidative damage in sicklered blood cells.

  4. Question 79 · Pharmacology

    The active metabolite of cyclophosphamide is:

    1. N - hydroxy cyclophosphamide
    2. 4 - hydroxy cyclophosphamide
    3. 4 - acetyl cyclophosphamide
    4. N - acetyl cyclophosphamide
    Show answer

    Answer: B. 4 - hydroxy cyclophosphamide

    The active metabolite of cyclophosphamide is 4 - hydroxy cyclophosphamide. Cyclophosphamide is an alkylating agent with antitumor and immunosuppressant properties. It isused for many solid tumors. The important side effects are cystitis due to metabolite acrolein andbladder carcinoma. Cystitis caused by cyclophosphamide is treated by systemic administration of mesna and byirrigating the bladder with acetylcysteine. Both these drugs contain –SH which detoxifies the toxicmetabolites. Plenty of fluids and frequent bladder voiding are also helpful.

  5. Question 80 · Pharmacology

    Nomogram used for vancomycin dosing is?

    1. Salazar
    2. Halsted
    3. Rummack-mathew
    4. Matzke
    Show answer

    Answer: D. Matzke

    The nomogram used for vancomycin dosing is the Matzke nomogram. The Matzke nomogram includes dosing intervals ranging from 0.5 to 12 days on a continuum ofCLcr(Creatinine clearance) values from 0 to 120 mL/min. It is designed to guide the determinationof the initial empiric dose and interval of intravenous (IV) vancomycin in hospitalized adults whoare not receiving hemodialysis. The nomogram does not apply to patients who exhibit highlyvariable vancomycin pharmacokinetics. These patients include the critically ill, those with burninjuries affecting more than 20 percent of their body surface, patients with rapidly changing renalfunction, pregnant women, those with liver failure/ascites, individuals at the extremes of weight,patients who are acutely post-transplant, and those with cystic fibrosis The Rumack-Matthew nomogram uses the serum acetaminophen concentration, in relation tothe time after ingestion, to assess potential hepatotoxicity after a single, acute ingestion ofacetaminophen.

  6. Question 81 · Pharmacology

    An anti-diabetic drug causes a decrease in HbA1C by causing an increase in GIP and GLP-1levels and a decrease in glucagon levels. It is an oral drug causing a reduction in both pre andpost-prandial glucose. The drug is:

    1. Pramlinitide
    2. Alogliptin
    3. Liraglutide
    4. Metformin
    Show answer

    Answer: B. Alogliptin

    Alogliptin is a dipeptidyl peptidase-4 (DPP) inhibitor. Glucagon-like peptide (GLP) andglucose-dependent insulinotropic polypeptide (GIP) are both incretins that promote the exocytosisof insulin into circulation. Both these incretins are rapidly inactivated by dipeptidyl peptidase. So,inhibition of dipeptidyl peptidase enhances the action of incretins and facilitates insulinrelease. Dipeptidyl peptidase-4 inhibitors enhance the release of insulin postprandially, decreaseglucagon, and thereby lower mealtime as well as fasting glucose. Other DPP-4 inhibitors include saxagliptin, linagliptin, alogliptin, and vildagliptin. They areweight neutral. DPP-4 inhibitors are associated with severe joint pain which reduces on discontinuation of thedrug. Linagliptin is cleared primarily by hepatobiliary system, with little renal clearance. Onlysaxagliptin is metabolized by the liver and must be used in lower doses if used along with acytochrome p450 enzyme inhibitor drug. Other drugs are renally excreted and hence must beused with caution in renal insufficiency. Other options: Option A - Pramlintide is an amylin agonist. It reduces postprandial glycemia and leads to weightloss. Option C - Liraglutide is also a GLP-1 receptor agonist and is given only in the subcutaneousinjectable form. It was recently approved for the treatment of obesity. Option D - Metformin belongs to the biguanides class of oral hypoglycemic agents. It acts byactivation of AMP-dependent protein kinase (AMPK). This results in reduced hepatic glucoseproduction. Side effects of metformin are lactic acidosis, gastrointestinal disturbance, and vitaminB12 deficiency.

  7. Question 82 · Pharmacology

    A patient is set to receive intermittent treatment with teriparatide. Which of the followingcomplications does this drug predispose him to?

    1. Osteonecrosis of the jaw bone
    2. Subtrochanteric fractures
    3. Net loss of bone mass
    4. Osteosarcoma
    Show answer

    Answer: D. Osteosarcoma

    The use of teriparatide predisposes an individual to the development of osteosarcoma. Teriparatide is a parathormone analog, which is given as subcutaneous injections.Pharmacokinetics and systemic actions of teriparatide on mineral metabolism are the same as forPTH. Its main target of action are the osteoblasts. Teriparatide is administered by once-daily subcutaneous injection of 20 µg into the thigh orabdomen. The site of administration should be rotated, but the time at which the injection ismade should be the same each day. Teriparatide and abaloparatide (PTHrP analog) are the only anabolic agents currently availablethat increase new bone formation (option C). Candidates for treatment include women who have ahistory of osteoporotic fracture, who have multiple risk factors for fracture, or who failed or areintolerant of previous osteoporosis therapy. Men with primary or hypogonadal osteoporosis arealso candidates for treatment with these agents. Adverse effects include hypercalcemia, exacerbation of nephrolithiasis, and elevation of serumuric acid levels. The development of osteosarcoma has been a serious concern in patients treatedwith teriparatide. It comes with a black-box warning, and use should be limited to no more than 2years and avoided in patients who are at increased baseline risk for osteosarcoma (including thosewith Paget disease of bone, unexplained elevations of alkaline phosphatase, open epiphyses, orprior radiation therapy involving the skeleton). Other options : Options A and B - Osteonecrosis of the jaw bone and subtrochanteric fractures are the adverseeffects of bisphosphonates.

Physiology

7 questions
  1. Question 83 · Physiology

    The formula for urea clearance is U*V/P, where U is

    1. Urine volume
    2. Serum urea in mg /100ml
    3. Urine urea in g/24h
    4. Urine urea in mg /100ml
    Show answer

    Answer: D. Urine urea in mg /100ml

    ‘U’ in the urea clearance stands for urine concentration of urea in mg /100ml. The formula for renal clearance of a substance is, Clearance (C) = U × V / P, where U = Urine concentration of the substance in mg / ml V = Urine flow rate in ml/min P = Plasma concentration of substance in mg /ml. Renal clearance of a substance is the volume of plasma that is completely cleared of the substanceby the kidneys per unit of time.

  2. Question 84 · Physiology

    The stroke volume depends on end-diastolic volume. This is best represented as

    1. Length-force relationship
    2. Length- velocity relationship
    3. Length-tension relationship
    4. Length-viscosity relationship
    Show answer

    Answer: C. Length-tension relationship

    The relationship between end-diastolic volume and stroke volume is best represented bylength-tension relationship. This is explained by the ‘Frank-Starling law of heart’ which states that ‘ the volume of bloodejected by the ventricle depends on the volume present in the ventricle at the end of the diastole’.The length of the myocytes depend on the volume (and hence tension) present at the end of thediastole, which in turn, depends on the volume returned to the heart, or the venous return. • The Frank-Starling law governs the normal ventricular function and ensures that the cardiacoutput equals venous return, within physiological limits. • There is a curvilinear relationship between stroke volume or cardiac output and the ventricularend-diastolic volume. • Agents that increase the contractility of the cardiac muscle have a positive inotropic effect andproduce increased stroke volume and thereby increased cardiac output for a given end-diastolicvolume. (Eg: Digoxin) • Agents that decrease the contractility of the cardiac muscle have a negative inotropic effect andproduce decreased stroke volume (and cardiac output) for a given end-diastolic volume.

  3. Question 85 · Physiology

    What does the zero pressure indicate in the pressure-volume curve?

    1. Residual volume
    2. Functional residual capacity
    3. Tidal volume
    4. Inspiratory reserve volume
    Show answer

    Answer: B. Functional residual capacity

    The zero pressure in the pressure-volume curve indicates the functional residual capacity. The pressure-volume relationship is basic to the understanding of the fundamental relationshipbetween the compliance of the lungs and the chest wall. It is measured by a spirometer. The functional residual capacity is the volume of the air that remains in the lungs after normalexpiration. The normal value is about 2.5 liters. Other options: Option A: Residual volume is the air left in the lungs after a maximal expiratory effort. Option C: Tidal volume is the amount of air that moves in during each inspiration or the amountof air that moves out with each expiration, during quiet breathing. The normal tidal volumeis 500–750 mL. Option D: Inspiratory reserve volume is the air inspired with a maximal inspiratory effort inaddition to the tidal volume.

  4. Question 86 · Physiology

    Fluid leaving the proximal convoluted tubule in the absence of antidiuretic hormone is:

    1. Isotonic
    2. Hypotonic
    3. Hypertonic
    4. Depends on fluid intake
    Show answer

    Answer: A. Isotonic

    Fluid leaving the proximal convoluted tubule (PCT) in the absence of antidiuretic hormone (ADH)is isotonic. The PCT is the site of maximum water reabsorption under normal physiological conditions.It reabsorbs around 60% of the filtered NaCl and water and other nutrients such as glucose andamino acids. Isosmotic reabsorption occurs in the PCT, so the amount of solute and waterreabsorbed are coupled and proportional to each other. Here, sodium is reabsorbed along with thewater. The large amounts of sodium and water reabsorption are essential in maintaining the ECFvolume. In addition to the isosmotic reabsorption, the water reabsorption in the PCT also occurs throughaquaporin-1 which are active water channels. Reabsorption of sodium and water in PCT is notaffected by ADH levels. Thus even in the absence of ADH, fluid is isotonic in PCT.

  5. Question 87 · Physiology

    Which of the following options is false regarding the measurement of temperature in a febrilepatient?

    1. Rectal temperature is ideal for measuring the core body temperature
    2. Rectal temperature is lower than axillary temperature
    3. Tympanic membrane temperature is lower than rectal temperature
    4. Rectal temperature is higher than oral temperature.
    Show answer

    Answer: B. Rectal temperature is lower than axillary temperature

    With regards to the measurement of temperature in a febrile patient, rectal temperature is higherthan axillary temperature. For humans, the conventional normal value for oral temperature is 37°C (98.6°F). Rectal temperature represents the core body temperature and is less affected by changes inenvironmental temperature. It remains relatively constant. In comparison, oral temperature istypically 0.5°C lower than rectal temperature. However, the oral temperature can be influenced byvarious factors such as consuming hot or cold fluids, chewing gum, smoking, and breathingthrough the mouth. Tympanic membrane thermometers measure radiant heat emitted by the tympanic membraneand the nearby ear canal, providing an absolute value of temperature. Tympanic membranereadings are generally 0.8°C (1.6°F) lower than rectal temperatures. The axillary temperaturemeasured is lower than the rectal temperature.

  6. Question 88 · Physiology

    Cells with prolific smooth endoplasmic reticulum secrete_______

    1. Eicosanoid hormones
    2. Steroid hormones
    3. Amine hormones
    4. Peptide hormones
    Show answer

    Answer: B. Steroid hormones

    Cells with prolific smooth endoplasmic reticulum secrete steroid hormones. The endoplasmic reticulum contains a complex series of tubules in the cytoplasm of the cell. • Ribosomes are attached to the cytoplasmic side of the membrane inthe rough(granular) endoplasmic reticulum. Protein synthesis and the initial folding ofpolypeptide chains with the formation of disulfide bonds are the functions of rough ER. • Ribosomes are absent in the smooth (agranular) endoplasmic reticulum. Free ribosomes arefound in the cytoplasm. It is the site of steroid synthesis in steroid-secreting cells and the site ofdetoxification processes in other cells. The sarcoplasmic reticulum is a modified endoplasmic reticulum, that functions asan intracellular calcium store, which releases calcium into the cytoplasm stimulating cellcontraction, proliferation, and migration.

  7. Question 89 · Physiology

    Which of the following does not stimulate catecholamine synthesis?

    1. Hyperglycemia
    2. Smoking
    3. Standing posture
    4. Exercise
    Show answer

    Answer: A. Hyperglycemia

    Hyperglycemia does not stimulate catecholamine synthesis. Instead, hypoglycemia stimulatescatecholamine synthesis. Catecholamine synthesis is primarily stimulated by various physiological stressors and factors thatactivate the sympathetic nervous system, such as physical exertion(exercise), changes in posture,or certain external stimuli like smoking. Other options: Option B: Smoking stimulates catecholamine release through the activation of the sympatheticnervous system, which increases catecholamines like adrenaline and norepinephrine. Option C: When a person stands up, it can lead to orthostatic hypotension, which triggers therelease of catecholamines to help increase blood pressure and maintain adequate perfusion. Option D: Exercise stimulates the sympathetic nervous system and increases catecholamineproduction to support increased heart rate, blood pressure, and metabolic demands.

Psychiatry

3 questions
  1. Question 90 · Psychiatry

    Which of the following is the most appropriate definition of stress?

    1. External pressure
    2. Mental tension
    3. It is the body's response to anything that requires attention or action causing physical,emotional, or psychological strain.
    4. Internal strain
    Show answer

    Answer: C. It is the body's response to anything that requires attention or action causing physical,emotional, or psychological strain.

    According to the WHO, stress can be defined as any type of change that causes physical,emotional, or psychological strain. Stress is your body's response to anything that requiresattention or action. Everyone experiences stress to some degree.

  2. Question 91 · Psychiatry

    SCOFF protocol is used for assessing:

    1. Mental health disorder
    2. Behavioural disorder
    3. Sexual health disorder
    4. Eating disorder
    Show answer

    Answer: D. Eating disorder

    The SCOFF (Sick, Control, One, Fat, Food) questionnaire is a screening tool for eatingdisorders (anorexia nervosa and bullimia nervosa). It consists of 5 questions addressing the core features of these disorders. The SCOFF questionsare: • Do you make yourself Sick because you feel uncomfortably full? • Do you worry that you have lost Control over how much you eat? • Have you recently lost more than One stone (~6.35 kg) in a 3-month period? • Do you believe yourself to be Fat when others say you are too thin? • Would you say that Food dominates your life? A 'Yes' answer to 2 or more questions out of the 5 indicates likely anorexia or bulimia.

  3. Question 92 · Psychiatry

    Which of the following is not true about 'rapid cyclers'?

    1. Occurs commonly in men
    2. Commonly seen with concomitant hypothyroidism
    3. At least 4 distinct episodes per year
    4. Can be triggered by antidepressants
    Show answer

    Answer: A. Occurs commonly in men

    Option A is the false statement. Rapid cycling disorder is more common in females. Rapid cycling disorder is diagnosed when the patient has a combination of 4 or more bipolarepisodes within the last 12 months. Two separate episodes can be demarcated from one anotherby either switch in polarity or full/partial remission for at least 2 months. Manic and hypomanicepisodes are considered to be on the same pole, hence a switch in polarity includes an episode ofmania/hypomania with an episode of depression. Possible risk factors for rapid cycling disorder include female patients, diabetes, childhood abuseor family trauma, and alcoholism or substance abuse. Rapid-cycling bipolar patients aresignificantly more likely than non-rapid-cycling patients to have a lifetime history of comorbiditieslike anxiety disorders, ADHD, OCD, bulimia nervosa, and substance use disorders. The drug of choice for prophylaxis of rapid cycling disorder is valproate Other options Option B: Conditions like hypothyroidism and diabetes are possibly involved in some cases ofrapid cycling . Option D: Usage of antidepressants could worsen rapid cycling in bipolar depression.

Radiology

1 question
  1. Question 93 · Radiology

    Egg shell calcification is seen in all the following, except _______.

    1. Silicosis
    2. Sarcoidosis
    3. Post-radiation therapy lymph nodes
    4. Metastatic lymph nodes
    Show answer

    Answer: D. Metastatic lymph nodes

    Eggshell calcification is not seen in the case of metastatic lymph nodes. A ring of calcification at the periphery of a lymph node is known as eggshell calcification.

Surgery

7 questions
  1. Question 94 · Surgery

    A patient who was admitted after a road traffic accident is put on mechanical ventilation. Heopens his eyes on verbal command and moves all four limbs spontaneously. Calculate hisGCS.

    1. Eyes-2, Verbal-1, Motor-5
    2. Eyes-2, Verbal-NT, Motor-5
    3. Eyes-3, Verbal-1, Motor-6
    4. Eyes-3, Verbal-NT, Motor-6
    Show answer

    Answer: D. Eyes-3, Verbal-NT, Motor-6

    The patients GCS is Eyes-3, Verbal-NT, Motor-6. It is calculated as follows: • Eye opening: Opening to verbal command - 3 • Verbal response: In the case of endotracheal intubation, the verbal response score is noted as NT(non-testable) • Motor respone: Movement of all 4 limbs spontaneously - 6 Total GCS score is not recorded since one of the components is non-testable.

  2. Question 95 · Surgery

    Frozen section was done during the surgery of carcinoma breast. This procedure will beuseful for all except

    1. To know sentinel lymph node
    2. To know margin positivity
    3. To know metastasis
    4. To confirm definitive diagnosis immediately in all cases
    Show answer

    Answer: D. To confirm definitive diagnosis immediately in all cases

    Frozen section analysis performed during breast carcinoma surgery is not suitable for definitivediagnosis due to technical challenges, potential quality issues in processing and staining, and thedifficulty of obtaining accurate results. Frozen section analysis is performed when a prompt biopsy report is required, allowing thesurgeon to make immediate decisions regarding the further steps of management. • Frozen section analysis plays a valuable role in assessing involvement of sentinel lymph nodesduring surgery (option A). When a positive node is identified intraoperatively, the surgeon canproceed with immediate axillary lymph node clearance. On the other hand, if the sentinel nodedoes not show tumor involvement, further axillary dissection can be avoided. • It also helps in diagnosing nodal metastasis in clinically identified lymph nodes (option C).However, it may not detect subcapsular micrometastases. • While performing breast-conserving surgeries, assessment of tumor margins (option B) usingfrozen sections is helpful in ensuring complete tumor excision.

  3. Question 96 · Surgery

    Which of the following would be the most commonly encountered ectopic tissue in Meckel'sdiverticulum?

    1. Gastric and pancreatic
    2. Colonic and pancreatic
    3. Gastric and colonic
    4. Thyroid and lung
    Show answer

    Answer: A. Gastric and pancreatic

    The ectopic tissue encountered in Meckel’s diverticulum includes gastric and pancreatic tissue. Meckel’s diverticulum is caused by a failure of normal regression of the vitelline duct thatdevelops during weeks 5 to 7 of gestation. The most common presentation is painless, massive, lower GI bleeds in children younger than 5years. A Meckel's can be the lead point for ileoileal or ileocolic intussusception. The bleedingMeckel's diverticulum may be confirmed by a 99 mTc-pertechnetate isotope scan to detect gastricmucosa and pancreatic tissue. Surgical treatment is the definitive therapy for Meckel’s diverticulum. A V-shapeddiverticulectomy with transverse closure of the ileum can be done.

  4. Question 97 · Surgery

    Which of the following is not a part of damage control surgery?

    1. Control of hemorrhage
    2. Resuscitation
    3. Decontamination
    4. Vascular anastomosis
    Show answer

    Answer: D. Vascular anastomosis

    Vascular anastomosis is undertaken as a part of definitive surgery, not damage controlsurgery (DCS). DCS is minimal intervention done in order to stabilize the patient and followed by definite surgerylater. It aims to coordinate staged operative interventions with periods of aggressiveresuscitation to salvage trauma patients sustaining major injuries. The initial focus is hemorrhage control, resuscitation, followed by control and limitation ofcontamination. It includes an abbreviated laparotomy, temporary packing, and closure of theabdomen in an eIort to blunt the physiologic response to prolonged shock and massivehemorrhage. After the initial steps of DCS, definitive surgery involving vascular anastomosis may be done. Thisis individualized to the patient and preferred to be completed in the first 24–72 hours. Note: Vascular anastomosis is the option least likely to be part of the initial stages of DCS. Hence,it is the best option out of the four for this testing point.

  5. Question 98 · Surgery

    In a patient who underwent laparoscopic cholecystectomy, what is the ideal urine output?

    1. 500 ml/day
    2. 0.5 ml/minute
    3. 0.5-1.5 ml/kg/hour
    4. 1 ml/kg
    Show answer

    Answer: C. 0.5-1.5 ml/kg/hour

    The ideal urine output in that patient following laparoscopic cholecystectomy is 0.5-1.5ml/kg/hour. Normal urine output is expected in the postoperative period of a patient who underwentlaparoscopic cholecystectomy. The average normal urine output in an adult is 0.5-1.5ml/kg/hour (Option B) or approximately 1500 mL/day (Option A). A minimum urine output of400 mL/day is essential to excrete the end products of protein metabolism. Option D: Normal urine output in ml/kg would be 0.5 - 1.5 ml/kg/hour multiplied by 24 whichis 12-36 ml/kg per day.

  6. Question 99 · Surgery

    What is the most common cause of hypercalcemia in hospitalized patients?

    1. Hypervitaminosis D
    2. Malignancy
    3. Parathyroid adenoma
    4. Parathyroid hyperplasia
    Show answer

    Answer: B. Malignancy

    Cancer is the most common cause of hypercalcemia in hospitalized patients, whereashyperparathyroidism is the most common cause of hypercalcemia detected incidentally inoutpatients. The causes of hypercalcemia in cancer : • Cancers affecting bone like multiple myeloma, leukemia, bony metastasis - cause the bone tobreak down which releases excess calcium into the blood • Pseudohyperparathyroidism - tumours producing parathyroid hormone-related peptide(PTHrP) cause the bone to release calcium into the blood. The most common carcinomaassociated is squamous cell carcinoma of lung. • Cancers affecting kidneys - failure to remove excess calcium from the blood • Dehydration caused by nausea and vomiting - difficult for the kidneys to remove calcium fromthe blood properly • Lack of physical activity - causes bone breakdown, releasing calcium into the blood

  7. Question 100 · Surgery

    What is the colour code in trauma triage for patients with life threatening conditions?

    1. Red
    2. Yellow
    3. Green
    4. Black
    Show answer

    Answer: A. Red

    Red is the colour code for patients with life threatening conditions who need immediate medicalcare. Triage includes rapid classification of the injured based on their injuries and the chances ofsurvival with prompt medical intervention. It provides maximum benefit to the greatest numberof injured in a disaster situation. It lessens the burden on medical facilities. The internationally accepted four-color coded system is used commonly. It includes: • Red - High priority attention or transfer; Critical, but likely to survive if treatment is given early • Yellow - Medium priority; Critical, likely to survive if treatment is given within hours • Green - Ambulatory patients; Stable, likely to survive even if treatment is delayed for hours todays • Black - Dead or moribund patients; Not breathing, pulseless, so severely injured that no medicalcare is likely to help

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