03Swt Bıochem

Test Name SWT-BIOCHEM & FMT 2017(MDMS) Total Questions 200

Test Type Examination Difficulty Level Difficult

Total Marks 600 Duration 180minutes

Test Question Language:- ENGLISH

(1). Embalming chemicals are a variety of preservatives, sanitising and disinfectant agents and additives used in modern embalming to temporarily prevent decomposition and restore a natural appearance for viewing a body after death Typically embalming fluid contains a mixture of

a. Formaldehyde, glycerine and ethanol

b. Formaldehyde, methanol and glycerine

c. Methanol,formaldehyde and ethanol

d. Glycerine, methanol and ethanol

Solution. (c) Methanol,formaldehyde and ethanol Ref– Read the text below Sol: - Embalming chemicals are a variety of preservatives, sanitising and disinfectant agents and additives used in modern embalming to temporarily prevent decomposition and restore a natural appearance for viewing a body after death. - A mixture of these chemicals is known as embalming fluid and is used to preserve deceased (dead) individuals, sometimes only until the funeral, other times indefinitely. - Typically embalming fluid contains a mixture of formaldehyde, methanol, ethanol and other solvents. - The formaldehyde content generally ranges from 5 to 29 percent and the ethanol content may range from 9 to 56 percent.

Correct Answer. c

(2). One of the following is true of antemortem abrasion

a. Bright red in colour

b. Exudation of serum is more

c. Scab formation

d. All of the above

Solution. (a) Bright red in colour Ref– Read the text below Sol: Difference Between Antemortem And Postmortem Bruise Antemortem abrasions will show signs of inflammation and repair whereas these will be absent in postmortem abrasions. The differences are-

Correct Answer. a

a. Premature babies.

b. Adults..

c. Children.

d. Old age.

Solution. (c) Children. Ref– Read the text below Sol: Growing fracture - In children, some linear fractures are associated with torn dura due to dense attachment of the dura to the bone in this age group. - A rare complication, which occurs in 0.6% of linear skull fractures in pediatric patients: 90% occur before the age of three. - The fracture is associated with a dural tear, preventing primary dural healing and resulting in progressive enlargement and eversion of

the fracture line. - The arachnoid sac may contain CSF and herniated brain with or without porencephaly.The treatment for the two forms is different, as the former requires a duro-cranioplasty (debridement of the damaged brain and dural repair) whilst the latter a shunt, in order to prevent progressive neurological deterioration.

Correct Answer. c

(4). Post mortem staining gets fixed after

a. 2-3 hours

b. 3-4 hours

c. 5-6 hours

d. 6-7 hours

Solution. (d) 6-7 hours Ref:Read the text below Sol - Fixation occurs earlier in summer and is delayed in asphyxial deaths.

- Some authors are of the opinion that hypostasis doesn’t get fixed. - Non displacement or non-shifting of lividity is due to haemoconcentration by loss of fluid which penetrates the walls of those vessels related to hydrostatic pressure

Correct Answer. d

(5). Overlapping of bones in intrauterine death is known as:

a. Spalding Sign

b. Robert’s Sign

c. Wood's Sign.

d. Ewald’s sign

Solution. (a) Spalding Sign Ref:Read the text below

Sol - Overlapping of bones in intrauterine death is known as Spalding Sign

Correct Answer. a

a. EMP pathway

b. HMP

c. TCA

d. Krebs

Solution. (b) HMP Ref:Read the text below Sol: - The pentose phosphate pathway (also called the phosphogluconate pathwayand the hexose monophosphate shunt) is a process that generates NADPH andpentoses (5-carbon sugars). - There are two distinct phases in the pathway. The first is the oxidative phase, in which NADPH is generated, and the second is the non- oxidative synthesis of 5-carbon sugars. This pathway is an alternative to glycolysis. - While it does involve oxidation of glucose, its primary role is anabolic rather than catabolic. For most organisms, it takes place in the cytosol; in plants, most steps take place in plastids

Correct Answer. b

(7). Lesch Nyhan syndrome is caused by-

a. HPRT complete deficiency

b. HPRT Partial deficiency

c. Purine nucleoside phosphorlase deficiency

d. PRP synthetase deficiency

Solution. (a) HPRT complete deficiency Ref:Read the text below Sol: - Mutations in the HPRT1 gene cause Lesch-Nyhan syndrome. - Mutations in the HPRT1 gene cause a severe deficiency of the enzyme hypoxanthine phosphoribosyltransferase 1.

- This enzyme is responsible for recycling purines, a type of building block of DNA and its chemical cousin RNA. - When this enzyme is lacking, the breakdown of purines results in abnormally high levels of uric acid in the body. - It is unclear how a shortage of this enzyme causes the neurological and behavioral problems characteristic of Lesch-Nyhan syndrome.

Correct Answer. a

(8). A polymorphism is best defined as

a. Cosegregation of alleles

b. One phenotype, multiple genotypes

c. One locus, multiple abnormal alleles

d. One locus, multiple normal alleles

Solution. (d) One locus, multiple normal alleles Ref– Read the text below Sol: - Polymorphic loci have multiple alleles because of DNA sequence variation, including one or more with frequencies greater than 1%. - This higher frequency and benign connotation differentiate polymorphic loci from those that harbor multiple disease-causing alleles. - The DNA sequence changes may alter restriction sites [producing restriction fragment length polymorphisms (RFLPs)], change the

numbers of repeated segments [producing variable numbers of tandem repeats (VNTRs)], or alter the genetic code (producing variant proteins, or protein polymorphisms). - Polymorphisms may cosegregate (be inherited together) with disease alleles, allowing diagnosis by linkage analysis or estimates of risk through allele association (a.k.a. linkage disequilibrium, as with certain HLA alleles and diabetes mellitus). - Different mutant alleles may cause indistinguishable phenotypes (allelic heterogeneity), as may mutations at different loci (genetic heterogeneity).

Correct Answer. d

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 4/86 (9). If a completely radioactive double-stranded DNA molecule undergoes two rounds of replication in a solution free of radioactive label, what is the radioactivity status of the resulting four double-stranded DNA molecules?

a. Half should contain no radioactivity

b. All should contain radioactivity

c. Half should contain radioactivity in both strands

d. One should contain radioactivity in both strands

Solution. (a) Half should contain no radioactivity Ref– Read the text below Sol: - The replication of double-stranded DNA is semiconservative, meaning that each strand separates and serves as a template for synthesis of a new complementary strand. - The first round of replication of a labeled DNA helix in a cold (unlabeled) solution will yield two daughter double-stranded molecules,

each with one labeled and one unlabeled strand. - The second round of replication will yield four double-stranded DNA molecules. - Two of these will have one original labeled strand and one unlabeled strand; the other two will have two unlabeled strands and contain no radioactivity.

Correct Answer. a

(10). Following ultraviolet damage of DNA in skin

a. A specific excinuclease detects damaged areas

b. Purine dimers are formed

c. Both strands are cleaved

d. Endonuclease removes the strand

Solution. (a) A specific excinuclease detects damaged areas Ref– Read the text below Sol: - Ultraviolet irradiation causes thymine dimers to form in DNA. Replication is inhibited in cells until the pyrimidine dimmers are removed. Removal of the damaged areas occurs in two ways. - The process can be simply reversed by a photoreactivating enzyme that cleaves the dimers and yields the original bases. Blue light is

required for this. - Alternatively, the dimer is removed. A UV-specific excinuclease nicks the dimer on its 5’side. DNA polymerase I replicates the damaged sequence,while the damaged sequence swings out. - Finally, the damaged piece is hydrolyzed by the 5′ to 3′ exonuclease activity of the DNA polymerase I. DNA ligase then joins the new piece to the original DNA at the cleavage site.

Correct Answer. a

(11). Charcoal therapy is indicated for the acute overdose ingestion of all the following medications, except:

a. Theophylline

b. A tricyclic antidepressant

c. Lithium

d. Aspirin

Solution. (c) Lithium Ref:Read the text below Sol -Charcoal therapy is indicated in the acute ingestion of all the drugs listed, except lithium. - Heavy metals are poorly bound by charcoal, and the risks outweigh the benefits of charcoal therapy in the acute setting.

Correct Answer. c

a. Soot in respiratory passages

b. Cyanosis of finger nails

c. Congestion of kidney

d. Pale internal organs

Solution. (a) Soot in respiratory passages Ref: Read the text below. Sol. - The presence of carbon particles in the terminal bronchioles on histological examination is absolute proof of life during fire.

- The soot is better seen by spreading a thin film of mucus on a clean sheet of white paper. - Presence & amount of soot in air-passage depends on the type of fire, the amount of smoke produced and duration of survival in smoke contaminated atmosphere.

Correct Answer. a

(13). Identify the fracture :

a. Communited fracture

b. Signature fracture

c. Pond’s fracture

d. Ring fracture

Solution. (c) Pond’s fracture Ref: Read the text below Sol: Other important facts to remember about pond fracture: (i) It is caused often by obstetric forceps (ii) Only outer table is fractured (iii) Resembles the indentation produced by squeezing a table tennis ball; when elevated it resumes and retains its normal position. Sometimes also known as ping-pong fracture for this reason. Variant: Pond fracture is common in (a)children (b)adults (c)elderly (d)Menopausal women. The answer is (a)children

Correct Answer. c

a. The eyes are bright, glistening and prominent with dilated pupils.

b. The jaws are firmly closed

c. Characteristic bitter lemon smell

d. The colour of the cheeks and postmortem staining may be cherry-red

Solution. (c) Characteristic bitter lemon smell Ref– Read the text below Sol: POST-MORTEM APPEARANCES OF CYANDINE POISONING - The colour of the cheeks and postmortem staining may be cherry-red in about half the cases, because oxygen remains in the cells as

oxyhaemoglobin, and due to the formation of cyanmethaemoglobin. - The odour of bitter almond may be noticed on opening the body. - The eyes may be bright, glistening and prominent with dilated pupils. - The jaws are firmly closed and there is froth at the mouth - The mucosa of the stomach may be eroded and blackened due to the formation of alkaline haematin.

Correct Answer. c

(15). Antemortem blister differs from postmortem blister by

a. Presence of Albumin & Chloride in blister fluid

b. Gas in blister

c. Dry hard surface of the floor of blister

d. Absence of hyperemia around the blister

Solution. (a) Presence of Albumin & Chloride in blister fluid Ref– Read the text below Sol: The antemortem blister shows no gas and hyperemia due to increased permeability of capillaries is prominent. Blisters are found mainly in scalds and heat burns. Antemortem blisters are mostly produced during burns.

Features of antemortem blisters: - Surrounded by an area of hyperemeia - Floor is reddened with swelling of papillae Features of postmortem blisters: - No surrounding hyperemeia - Floor is not reddened

Correct Answer. a

(16). Which of the following enzymes can polymerize deoxyribonucleotides into DNA?

a. DNA ligase

b. DNA gyrase

c. RNA polymerase III

d. Reverse transcriptase

Solution. (d) Reverse transcriptase Ref– Read the text below Sol: - Reverse transcriptase is an RNA-dependent DNA polymerase that can synthesize first a single strand and then a doublestranded DNA from a single-strand RNA template.

- It was originally found in animal retroviruses. Primase is a DNA-dependent RNA polymerase enzyme that synthesizes an RNA molecule 10 to 200 nucleotides in length that initiates or “primes” DNA synthesis. - DNA ligase joins DNA fragments and DNA gyrase winds or unwinds DNA. - Transfer RNA, 5SRNA, and other small RNAs are synthesized by RNA polymerase III (RNA polymerase I synthesizes ribosomal RNA and RNA polymerase II synthesizes messenger RNA).

Correct Answer. d

(17). Which of the following molecules is found in a nucleoside?

a. A pyrophosphate group

b. A 1′-base linked to a pentose sugar

c. A 5′-phosphate group linked to a pentose sugar

d. A 3′-phosphate group linked to a pentose sugar

Solution. (b) A 1′-base linked to a pentose sugar Ref– Read the text below Sol: - A nucleoside consists of a purine or pyrimidine base linked to a pentose sugar. - The 1′-carbon of the pentose is linked to the nitrogen of the base. In DNA, 2′-deoxyribose sugars are used; in RNA, ribose sugars are used. - Nucleotides are phosphate esters of nucleosides with one to three phosphate groups, such as adenosine monophosphate (AMP), adenosine diphosphate (ADP), or adenosine triphosphate (ATP). - The nitrogenous bases are adenine, thymine, guanine, and cytosine in DNA, with thymine replaced by uridine in RNA. Nucleotide polymers are chains of nucleotides with single phosphate groups, joined by bonds between the 3′-hydroxyl of the preceding pentose and the 5′-phosphate of the next pentose. - Polymerization requires high-energy nucleotide triphosphate precursors that liberate pyrophosphate (broken down to phosphate) during joining. - The polymerization reaction is given specificity by complementary RNA or DNA templates and rapidity by enzyme catalysts called polymerases.

Correct Answer. b

(18). Which is the most correct sequence of events in gene repair mechanisms in patients without a mutated repair process?

a. Nicking, excision, replacement, sealing, recognition

b. Sealing, recognition, nicking, excision, replacement

c. Recognition, nicking, excision, replacement, sealing

d. Nicking, sealing, recognition, excision, replacement

Solution. (c) Recognition, nicking, excision, replacement, sealing Ref– Read the text below Sol: In all of the forms of DNA repair in normal cells, a common sequence of events occurs. 1. The single or multiple base abnormality is surveyed and detected by a specific protein or proteins. 2. The DNA is nicked on one side of the damaged DNA. 3. A specific enzyme excises the damaged portion (steps 2 and 3 can be combined if an excinuclease cuts on both sides of the damaged DNA). 4. The damaged portion of the strand is replaced by resynthesis catalyzed by DNA polymerase I. 5. A ligase seals the final gap. With some variability, these general principles apply in nucleotide excision repair (segments of about 30 nucleotides), base excision repair of single bases, and mismatch repair of copying errors (one to five bases).

Correct Answer. c

a. Stop codons

b. Peptidyl transferase

c. Release factors

d. Dissociation of ribosomes

Solution. (b) Peptidyl transferase Ref– Read the text below Sol: - During the course of protein synthesis on a ribosome, peptidyl transferase catalyzes the formation of peptide bonds. However, when a stop codon such as UAA, UGA, or UAG is reached,aminoacyl-tRNA does not bind to the A site of a ribosome. - One of the proteins, known as a release factor, binds to the specific trinucleotide sequence present.

- This binding of the release factor activates peptidyl transferase to hydrolyze the bond between the polypeptide and the tRNA occupying the P site. - Thus, instead of forming a peptide bond, peptidyl transferase catalyzes the hydrolytic step that leads to the release of newly synthesized proteins. - Following release of the polypeptide, the ribosome dissociates into its major subunits.

Correct Answer. b

(20). The function of signal recognition particles is to

a. Cleave signal sequences

b. Detect cytosolic proteins

c. Direct the signal sequences to ribosomes

d. Bind ribosomes to endoplasmic reticulum

Solution. (d) Bind ribosomes to endoplasmic reticulum Ref– Read the text below Sol: - The directing of nascent polypeptide chains to the endoplasmic reticulum is regulated by signal recognition particles (SRPs). - The signal sequence of a nascent protein is recognized by an SRP, which complexes with the ribosome, mRNA, and the nascent protein. - The complexed SRP then binds to an SRP receptor on the surface of the endoplasmic reticulum. After the ribosome is transferred to ribophorins and the translocation begins, SRP is released back into the cytosol. - Ribosomes with nascent protein without a signal sequence do not participate in this process and instead synthesize proteins that are released into the cytosol.

Correct Answer. d

(21). In a charred body which of the following means is useful it its identification

a. Stature

b. Comparison of post mortem x rays with dental records

c. Scar marks

d. Skeletal features

Solution. (b) Comparison of post mortem x rays with dental records Ref– Read the text below

Sol: - Comparison of post mortem x rays with dental records is recommended for the identification of charred body

Correct Answer. b

a. Salivary dribbling

b. Congestion of lungs

c. Ligature marks

d. Petechial hemorrhages

Solution. (a) Salivary dribbling Ref– Read the text below Sol: Hanging remains to be one of the common methods of committing suicide. Homicidal and accidental hanging is rare. Hence all cases of hanging are considered suicidal until the contrary is proved. Because of the above, postmortem suspension of the body may be resorted to cloak a crime. Therefore meticulous dissection and sharp distinction between ante mortem hanging and postmortem suspension is warranted. There is no specific gold standard to distinguish between ante mortem hanging and postmortem suspension. How ever presence of- - Vertical salivary dribble mark from the dependant angle of mouth. - The phenomenon of Le Facies sympathique. - Hyperemia and echymosis of margins of ligature mark. - Horizontal tear of the intima of carotid artery at level of ligature with infiltration around. are considered as ante mortem features of hanging. But obvious salivary dribble mark could be detected only in 56% of cases

Correct Answer. a

(23). Motile spermatozoa found in wet mount of vaginal secretions are indicative of intercourse within the past :

a. 6 hours

b. 12 hours

c. 24 hours

d. 48 hours

Solution. (c) 24 hours Ref– Read the text below

Sol: Research had proved that following sexual intercourse motile spermatozoa should be found in vagina for 24 hrs,that they are likely to be found upto 3 days later,and occasionally they are found 7 days later.

Correct Answer. c

a. Accidental injury

b. Suicidal wounds

c. Homicidal wounds

d. None of the above

Solution. (b) Suicidal wounds Ref– Read the text below Sol: - Hesitation, or tentative, wounds are defined either as: any cut or wound that is self inflicted after a decision is made not to commit suicide, or any tentative cut or wound that is made before the final cut that causes death. - Such wounds are usually superficial, sharp, forced skin cuts found on the body of victims. These less severe cutting marks are often caused by attempts to build up courage before attempting the final, fatal wound. - Non-fatal, shallow hesitation wounds can also accompany the deeper, sometimes fatal incisions. - Although hesitation cuts are not always present in cases of suicide, they are typical of suicidal injuries. However, the presence of hesitation marks alongside or near to the final fatal mark usually indicates a forensic diagnosis of suicide over other possible causes of death. - Hesitation wounds are generally straight-line marks at the elbows, neck/throat, and wrists, although in a few cases they occur in the general area of the upper middle part of the abdomen (near the heart). - Wounds made by people attempting suicide are typically made at an angle related to the hand that holds the weapon. The angle of such hesitation wounds is usually in a downward flowing direction because of the natural motion of the arm as it sweeps across the body. - Hesitation wounds are often made under clothing, with particular parts of the clothing being parted to expose the target area of the body, a common feature seen by forensic experts examining suicidal wounds.

Correct Answer. b

(25). Which of the following is an early sign of death ?

a. Rigor mortis

b. Adipocere formation

c. Putrification

d. Mummification

Solution. (a) Rigor mortis Ref– Read the text below Sol: - Rigormortis is a condition characterised by stiffening, shortening and opacity of the muscles which follow the period of primary relaxation. - It is due to chemical changes involving the proteins of the muscle fibres and it marks the end of the muscles, cellular or molecular life. - In India, rigormortis commences in 1-2 hours after death, takes about 12 hours to develop from head to foot, persists for another 12 hours,and takes 12 hours to pass off. - Thus, the presence and extent, or absence of rigormortis helps to provide a right estimation of the time since death. - With the disappearance of rigormortis, the muscles becomes soft and flaccid once again but do not respond to mechanical or electrical stimuli.

Correct Answer. a

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 11/86 (26). A 25-year-old male gets into a brawl outside a bar. During the altercation, someone pulls out a gun and shoots him in the head.The bullet enters the man's temple and severs his right optic nerve completely. He is quickly transported to a nearby emergency room and an emergency physician tests his pupillary response by shining a light in the right eye. What will the physician most likely find?

a. No pupillary constriction in the right eye, and no pupillary constriction in the left eye

b. No pupillary constriction in the right eye, but pupillary constriction in the left eye

c. Pupillary constriction followed by pupillary dilatation in both eyes

d. Pupillary constriction in the right eye, and no pupillary constriction in the left eye

Solution. (b) No pupillary constriction in the right eye, and no pupillary constriction in the left eye. Ref:Read the text below Sol: - This person is blind in the right eye. - The afferent limb of the pupillary light reflex is carried by the optic nerve (CN II), and the efferent limb is via the oculomotor nerve (CN III), which carries parasympathetic fibers from the Edinger-Westphal nucleus. - Thus, shining a light in the affected eye will not elicit any pupillary response. - On the other hand, shining the light in the left eye will result in simultaneous constriction of both pupils (assuming an intact right CN III), since the left optic nerve is intact.

Correct Answer. b

(27). Medical qualifications awarded by institutions out side India and recognized by MCI are registered in:

a. First schedule of Indian medical Council Act 1956.

b. Second schedule of Indian Medical Council Act 1956.

c. Part I of third schedule of Indian Medical Council Act 1956.

d. Part II of third schedule of Indian Medical Council Act 1956.

Solution. (d) Part II of Third schedule of Indian Medical Council Act. Ref:Read the text below Sol: RECOGNITION OF FOREIGN MEDICAL QUALIFICATIONS - If an Indian national obtains a foreign qualification which is not included in part II of the Third Scheduled, he can apply to the Central Government. - The candidate is required to provide full information with regard to the course of study, syllabus duration of the course etc. - The Central Government may, by notification in the Official Gazette, amend the part II of the Third Schedule so as to include such qualification therein.

Correct Answer. d

(28). Which of the following best describes the degree of extension of fourth-degree burns?

a. Into the reticular layer of the dermis

b. Into the papillary layer of the dermis

c. Into the epidermis

d. Into the subcutaneous fat, muscle,and bone

Solution. (d) Into the subcutaneous fat, muscle,and bone. Ref:Read the text below Sol: - A first-degree burn involves the epidermis. Second-degree burns may be superficial (papillary layer) partial thickness and deep (reticular layer) partial thickness. - Third-degree burns are full-thickness burns that involve the entire thickness of the skin. - Fourth-degree burns extend through the skin to subcutaneous fat, muscle,and bone. - The rule of 9’s is often used to calculate burn surface area in adults: 9% for each arm and the head and 18% for each leg and each side of the torso.

Correct Answer. d

a. 3 years

b. 4 years

c. 5 years

d. 6 years

Solution. (b) 4 years Ref: Read the text below Sol: Age of appearance of carpal bones Capitate: 2 months Hamate: 3 months

Lunate 3 years Scaphoid: 4 years Trapezium: 4-5 years Trapezoid: 4-5 years Triquetrum: 4-5 years Pisiform: 9-11 years

Correct Answer. b

(30). A noncompetitive inhibitor of an enzyme

a. Increases Km with no or little change in Vmax

b. Decreases Km and decreases Vmax

c. Decreases Vmax

d. Increases Vmax

Solution. (c) Decreases Vmax Ref– Read the text below Sol: - In contrast to competitive inhibitors, noncompetitive inhibitors are not structural analogues of the substrate. - Consequently, noncompetitive inhibitors bind to enzymes in locations remote from the active site. - For this reason, the degree of inhibition is based solely upon the concentration of inhibitor and increasing the substrate concentrations do not compete with or change the inhibition. - Therefore, unlike the increase in Km seen with competitive inhibition, in noncompetitive inhibition Vmax increases while Km usually remains the same. - While competitive inhibitors can be overcome at sufficiently high concentration of substrate, noncompetitive inhibition is irreversible

Correct Answer. c

a. Effectors may enhance or inhibit substrate binding

b. They are not usually controlled by feedback inhibition

c. The regulatory site may be the catalytic site

d. Michaelis-Menten kinetics describe their activity

Solution. (a) Effectors may enhance or inhibit substrate binding Ref– Read the text below Sol: - The binding of an effector to the regulatory subunit of an allosteric enzyme causes a conformational change that either increases or decreases the activity of the enzyme’s separate catalytic site. - Only in some allosteric molecules, such as hemoglobin, does positive cooperativity occur. A positive effector increases substrate binding. - This is the case with cyclic AMP dependent protein kinase of the glycogen phosphorylase cascade. Cyclic AMP binds the regulatory subunit that dissociates from the catalytic subunit and thereby activates it. In the absence of cyclic AMP, the regulatory subunit tightly binds the catalytic subunit and inactivates the enzymes. - Many allosteric enzymes are often placed at the first, or committed, step of a metabolic pathway. The end product of the pathway then acts as a negative effector of the enzyme. This is called feedback inhibition. - An allosteric enzyme does not obey Michaelis-Menten kinetics.

Correct Answer. a

(32). Which one of the following enzymes catalyzes high-energy phosphorylation of substrates during glycolysis?

a. Pyruvate kinase

b. Phosphoglycerate kinase

c. Triose phosphate isomerase

d. Glyceraldehyde-3-phosphate dehydrogenase

Solution. (d) Glyceraldehyde-3-phosphate dehydrogenase Ref– Read the text below Sol: - High-energy phosphate bonds are added to the substrates of glycolysis at three steps in the pathway.Hexokinase—or, in the case of liver, glucokinase—adds phosphate from ATP to glucose to form glucose- 6-phosphate. - Strictly speaking, this is not always considered a step of the glycolytic pathway. Phosphofructokinase uses ATP to convert fructose-6-phosphate to fructose-1,6-phosphate. - Using NAD+ in an oxidation-reduction reaction, inorganic phosphate is added to glyceraldehyde-3-phosphate by the enzyme glyceraldehyde-3- phosphate dehydrogenase to form 1,3-diphosphoglycerate. - The enzymes phosphoglycerate kinase and pyruvate kinase transfer substrate highenergy phosphate groups to ADP to form ATP.

Correct Answer. d

a. Hexokinase

b. Phosphofructokinase

c. Glyceraldehyde-3-phosphate dehydrogenase

d. Phosphoglycerate kinase

Solution. (c) Glyceraldehyde-3-phosphate dehydrogenase Ref– Read the text below Sol: - All the enzymes named are glycolytic enzymes that carry out phosphorylation of glucose-derived substrates or of ADP to form ATP. - However, only the reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase is a phosphorylation reaction coupled to oxidation that uses inorganic phosphate.

- In this reaction, glyceraldehyde-3-phosphate is converted to 1,3 bisphosphoglycerate by the addition of inorganic phosphate and the oxidation of glyceraldehyde- 3-phosphate with the concomitant reduction of NAD+ to NADH + H+. - This reaction is an example of a high-energy phosphate compound being produced by an oxidation-reduction reaction. The oxidation of the aldehyde group at C1 of glyceraldehyde-3-phosphate provides the energy for the reaction. - The 1,3-bisphosphoglycerate can then be utilized to phosphorylate ADP to ATP through the action of phosphoglycerate kinase, which is the next step in the glycolytic pathway

Correct Answer. c

(34). Which of the following hormones stimulates gluconeogenesis?

a. Progesterone

b. Glucagon

c. Aldosterone

d. Epinephrine

Solution. (b) Glucagon Ref– Rea d the text below Sol: - The balance and integration of the metabolism of fats and carbohydrates are mediated by the hormones insulin, glucagon, epinephrine, and norepinephrine. - All of these hormones exercise acute effects upon metabolism. Glucagon stimulates gluconeogenesis and blocks glycolysis. - When blood sugar levels get low, the α cells of the pancreas release glucagon. The main targets of glucagon are the liver and adipose tissue. - In the liver, glucagon stimulates the cyclic AMP–mediated cascade that causes phosphorylation of phosphorylase and glycogen synthesis. - This effectively turns off glycogen synthase and turns on glycogen phosphorylase, thereby causing a breakdown of glycogen and a production of glucose in liver, which ultimately raises blood glucose levels. - Insulin and glucagon are two antagonistic hormones that maintain the balance of sugar and fatty acids in blood. Insulin is produced by the - cells of the pancreas and its release is stimulated by high levels of glucose in the blood. It has a number of effects, but its major effect is to allow the entry of glucose into cells. - Insulin also allows the dephosphorylation of key regulatory enzymes. The consequence of these actions is to allow glycogen synthesis and storage in both muscle and liver, suppression of gluconeogenesis, acceleration of glycolysis, promotion of the synthesis of fatty acids, and promotion of the uptake and synthesis of amino acids into protein. - All in all, insulin acts to promote anabolism.

Correct Answer. b

a. Reduced nicotinamide dinucleotide (NADH)

b. Adenosine diphosphate (ADP)

c. Guanosine triphosphate (GTP)

d. Nicotinamide dinucleotide phosphate (NADP+)

Solution. (d) Nicotinamide dinucleotide phosphate (NADP+) Ref– Read the text below Sol: - The pentose phosphate pathway (hexose monophosphate shunt) functions to generate NADPH for reductive synthesis of compounds such as fatty acids or steroids, and to generate ribose for nucleotide and nucleic acid synthesis.

Correct Answer. d

(36). In Type I diabetes, the increased production of ketone bodies is primarily a result of which of the following?

a. A substantially increased rate of fatty acid oxidation by hepatocytes

b. An increase in the rate of the citric acid cycle

c. Decreased cyclic adenosine monophosphate (cAMP) levels in adipocytes

d. Elevated acetyl-coa levels in skeletal muscle

Solution. (d) elevated acetyl-CoA levels in skeletal muscle Ref:Read the text below Sol: - In fasting or diabetes, lipolysis predominates in adipocytes because of the inability of these cells to obtain glucose, which is normally used as a source of glycerol-3-phosphate. Glycerol-3-phosphate is necessary for the esterification of fatty acids into triacylglycerides. -Circulating fatty acids become the predominant fuel source, and beta-oxidation in the liver becomes substantially elevated. - This leads to an increased production of acetyl-CoA. Although gluconeogenesis is increased in the liver as a result of the persistent

elevation of glucagon levels, this pathway does not supply acetyl-CoA for the production of ketone bodies. The increased gluconeogenesis predisposes oxaloacetate and reduces the flow of acetyl-CoA through the citric acid cycle. - As a consequence, acetyl-CoA is diverted to the formation of ketone bodies. The persistently elevated levels of glucagon also increase the levels of cAMP in responsive tissues, such as adipocytes . This effect in adipocytes leads to persistently increased release of fatty acids to the circulation. - Since skeletal muscle lacks receptors for glucagon, there is no diabetes-mediated increase in muscle metabolism and thus no elevation in acetyl-CoA levels in skeletal muscle

Correct Answer. d

(37). Which of the following symptoms can occur frequently in infants suffering from mediumchain acyl-CoA dehydrogenase (MCAD) deficiency if periods between meals are protracted?

a. Bone and joint pain and thrombocytopenia

b. Hyperammonemia with decreased ketones

c. Hyperuricemia and darkening of the urine

d. Hypoglycemia and metabolic acidosis with normal levels of ketones

Solution. (d) Hypoglycemia and metabolic acidosis with normal levels of ketones Ref:Read the text below Sol: - In infants, the supply of glycogen lasts less than 6 hours and gluconeogenesis is not sufficient to maintain adequate blood glucose levels. - Normally, during periods of fasting (in particular during the night) the oxidation of fatty acids provides the necessary ATP to fuel hepatic gluconeogenesis as well as ketone bodies for nonhepatic tissue energy production. - In patients with MCAD deficiency there is a drastically reduced capacity to oxidize fatty acids. This leads to an increase in glucose usage with concomitant hypoglycemia. The deficit in the energy production from fatty acid oxidation, necessary for the liver to use other carbon sources, such as glycerol and amino acids, for gluconeogenesis further exacerbates the hypoglycemia. - Normally, hypoglycemia is accompanied by an increase in ketone formation from the increased oxidation of fatty acids. In MCAD deficiency there is a reduced level of fatty acid oxidation, hence near normal levels of ketones are detected in the serum.

Correct Answer. d

(38). Action of following enzyme leads to formation of superoxide ions

a. Superoxide dismutase

b. Catalase

c. L-amino acid oxidase

d. Glutathione peroxides

Solution. (c) L-amino acid oxidase Ref:Read the text below Sol: When oxygen molecule takes up one electron, by univalent reduction, it becomes superoxide anion ‘O2-‘ Superoxide anion is formed by: - xanthine oxidase - aldehyde dehydrogenase - L-amino oxidase - univalent oxidation wih molecular oxygen in the respiratory chain - during methaemoglobin formation

- during cytosolic hyroxylations of steroids, drugs etc by CYP450 or CYP448 systems - ionizing radiations - during phagocyosis Free radicals are scavenged by - Superoxide dismutase - Catalase - Glutathione peroxidase - Ferricytochrome - Endogenous ceruloplasmin

Correct Answer. c

(39). A red cell will swell when it is placed in a solution of

a. 100m Molar calcium chloride

b. 150m Molar sodium chloride

c. 200m Molar potassium chlorie

d. 250m Molar urea

Solution. (d) 250m Molar urea Ref:Read the text below Sol: - If a red cell is placed in a hypotonic solution, the cell will swell until intracellular osmolarity is same as extracellular osmolarity.

- The tonicity of extracellular solution is equal to the osmolarity of those particles w ithin the extracellular solution that are not permeable to the cell membrane. - Because urea is able to permeate the cell membrane, its effective osmotic pressure is zero. Therefore, RBC’ s placed in urea will swell until hemolysis.

Correct Answer. d

a. β–hydroxy-β-methyl glutaryl COA →Mevalonic acid

b. Acetyl COA + CO2 + ATP → Malonyl COA + ADP + Pi

c. L-Methyl malonyl COA → Succinyl COA

d. Lactose + H2O → Glucose + Galactose

Solution. (a) β–hydroxy-β-methyl glutaryl COA →Mevalonic acid Ref.:Read the text below Sol : - The enzyme “HMG – COA REDUCTASE” is required in the 3rd step of cholesterol biosynthesis. - This is the ‘rate limiting step’ in cholesterol biosynthesis. - It is a cytoplasmic enzyme. - It uses 2 molecules of NADPH

Correct Answer. a

(41). In the diagram below, which of the following conditions is most likely to shift the oxyhemoglobin curve from B to A?

a. Hypothermia

b. Hyperventilation

c. Carbon monoxide poisoning

d. Exercise

Solution. (d) Exercise Ref– Read the text below Sol: - During exercise, there is an increase in muscle temperature, carbon dioxide, and H-+, all of which shift the oxyhemoglobin curve to the right and enhance oxygen release to the tissues. - Hypothermia, hyperventilation (hypocapnic alkalosis), carbon monoxide, and transfusion with banked blood (decreased 2,3- bisphosphoglycerate) all shift the oxyhemoglobin dissociation to the left.

Correct Answer. d

a. Di hydro uracil

b. Adenine

c. Cytosine

d. Uracil

Solution. (a) Di hydro uracil Ref.: Read the text below Sol : Two types of nitrogenous bases are present in all nucleic acids namely : Purines : - Adenine - Guanine Pyrimidines : - Cytosine - Thymine - Uracil A few other modified pyrimidine bases like dihydrouracil and 5-methyl cytosine are rarely found in some types of RNA.

Correct Answer. a

(43). A 51-year-old homeless man presents to the emergency room in wintertime complaining of numbness of his feet. Physical examination of the feet reveals erythema, edema, and the presence of several clear blisters. Peripheral pulses are palpable. Which of the following is the most likely diagnosis?

a. Frostnip

b. First-degree frostbite injury

c. Second-degree frostbite injury

d. Third-degree frostbite injury

Solution. (c) Second-degree frostbite injury. Ref:Read the text below Sol: - Frostnip is a superficial freeze injury that causes no tissue loss. Patients complain of some discomfort and the involved area is pale; rewarming quickly reverses the symptoms. First-degree frostbite is characterized by partial skin freezing,erythema, edema, no blisters, and desquamation several days later. - Second-degree frostbite is characterized by full-thickness skin freezing,erythema, edema, and the presence of clear blisters. Patients

complain of throbbing and numbness. - Third-degree frostbite injuries are characterized by damage that extends into the subdermal plexis. The skin is blue or gray and there are hemorrhagic blisters. Patients complain of burning,shooting pains, and the feeling that the involved area feels like “a block of wood.” Prognosis is poor. - Fourth-degree frostbite injuries extend into the subcutaneous tissue, muscle, and bone. There is typically no edema and the skin is mottled and cyanotic; eventually these injuries form a mummified eschar.

Correct Answer. c

a. Inverted edges and linear

b. Inverted edges and circular

c. Everted edges and linear

d. Everted edges and circular

Solution. (b) Inverted edges and circular Sol: CLOSE SHOT - This term is applied when the victim is within the range of the flame, i.e. 5 to 8 cm. The term ‘point blank’ is used when the range is

very close to or in contact with the surface of the skin. - The entrance wound is circular with inverted edges, but the rebounding gases may level up or even evert the margins. - The skin surrounding the wound is hyperaemic and shows some bruising, burning, blackening and tattooing. - Carboxy-haemoglobin will be present in the wound track in diminishing concentrations.

Correct Answer. b

(45). “Back spatter” is related with by :

a. Inverted edges and linear

b. Inverted edges and circular

c. Contact shot

d. Close shot

Solution. (c) Contact shot Ref: Read the text below Sol: CONTACT SHOT: - The wound is large shows cavitation, and triangular stellate, cruciate or elliptic. - The margins are contused and everted due to gases coming out of the entering wound under pressure. - There is no burning blackening and tattooing around the wound entrance. Tattooing is minimal or absent. The wound is not eruptive or explosive in appearance. - In contact shot, the muzzle blast and the negative pressure in the barrel following discharge may suck blood, hair, fragments of tissues and cloth fibers several cm. back inside the barrel called “back spatter”. - The corona consists of a circular zone of soot deposit surrounding the bullet defect, but separated from it by a band of skin without a deposit of soot. - Singeing of the hair may also be seen.

Correct Answer. c

(46). Choose the most likely associated odor with diabetic coma :

a. Fruity odor

b. Bitter almond odor

c. Burned-rope odor

d. Rotten eggs odor

Solution. (a) Fruity odor Ref: Read the text below Sol: - Cyanide poisoning is associated with a bitter almond odor and diabetic acidosis is associated with a fruity odor. - Arsenic ingestion and parathion poisoning are associated with a garlic odor.

Correct Answer. a

a. Phenytoin.

b. Methyl alcohol.

c. Lithium

d. Ethylene glycol

Solution. (a) Phenytoin. Ref: Read the text below Sol: - Hemoperfusion should be considered in cases of severe poisoning due to

Correct Answer. a

(48). Frequent blushing (“erethism”) is associated with : -

a. Mercury

b. Lead

c. Phenolic acid

d. Carbolic acid

Solution. (a) Mercury Reference – Poisoning & Drug Overdose by Kent R. Olson - 213 Ans Mercury Mechanism of toxicity. Mercury reacts with sulfhydryl (SH) groups, resulting in enzyme inhibition and pathologic alteration of cellular membranes. Clinical presentation. Acute inhalation - Severe chemical pneumonitis and noncardiogenic pulmonary edema. - Acute gingivostomatitis may also occur. Chronic intoxication Classic triad - Tremor - Neuropsychiatric disturbances - Gingivostomatitis. Metallic taste  Frequent blushing (“erethism”).  Pain in the extremities,often accompanied by pinkish discoloration and desquamation (“pink disease”)

Correct Answer. a

a. Ethanol

b. Phenol

c. Glycerine

d. Formalin

Solution. (a) Ethanol References: 1. Pg. 613, 803, 1535; Dorland’s Medical dictionary (31st Ed.) 2. Pg. 112; Embalming principles – Dr Ajmani (1st Ed.) 3. Textbook of forensic medicine & toxicology by RK Sharma 150- 151,parikh, 4. Reddy Pg 153 27th ed Sol: - Ethanol is a preservative and can be used for embalming but is not the usual content of Embalming fluid/solution. - Instead Methanol is used, which is cheaper and more toxic to bacteria than ethanol. Emblaming defined as – study and science of treating dead body to achieve antiseptic condition, a life-like postmortem appearance and preservation. Uses – bodies in M.C. for

dissection, transport of dead bodies Principle-- coagulation of protein, tissue & fixed organs are bleached and hardened Chemicals MC used – formaldehyde – dehydrate, hardens - Formalin – 60% - Methanol – 25% - Liquefied phenol – 10% - Sodium lauryl sulphate– 1% - Mercuric chloride – 1% - Eucalyptus oil– 1%

Correct Answer. a

a. Dolicocephalic skull

b. Brachycephalic skull

c. Trigonocephalic skull

d. Plagiocephalic skull

Solution. (b) Brachycephalic skull Ref: Read the text below Sol:

Cephalic index = (maximum breadth of the skull/ maximum length of the skull) X 100. Dolicocephalic skull – CI – 70 – 75 Brachycephalic skull – CI – 80 – 85 Plagiocephaly, also known as flat head syndrome,is a condition characterized by an asymmetrical distortion (flattening of one side) of the skull. It is characterized by a flat spot on the back or one side of the head caused by remaining in a particular position for too long Trigonocephaly (Greek: 'trigonon' = triangle, 'kephale' = head) is a congenital condition of premature fusion of the metopic suture (Greek: 'metopon' = forehead) leading to a triangular shaped forehead. The merging of the two frontal bones leads to transverse growth restriction and parallel growth expansion. It may occur syndromic involving other abnormalities or isolated.

Correct Answer. b

(51). For biochemical analysis vitreous in sent in:

a. Hydrochloric acid

b. Phenol

c. Formalin

d. Fluoride.

Solution. (d) Fluoride. References: Read the text below Sol: Potassium/sodium fluoride: - Collecting vitreous humour sample for ethanol analysis in postmortem toxicology cases. A preservation(sodium fluoride,0.5-2% weight by volume (w/v) should be added.

Correct Answer. d

a. 100 mg/dL

b. 200 mg/dL

c. 300 mg/dL

d. >400 mg/dL

Solution. (d) >400 mg/dL. Ref– Read the text below Sol: - Behavioral changes, slowing of motor performance, and decrease in the ability to think clearly may appear with a blood alcohol level as low as 20–30 mg/dL. - Most people show significant impairment of motor and mental performance when their alcohol levels reach 100 mg/dL.

- With blood alcohol concentration between 200 and 300, slurred speech is more intense and memory impairment, such as blackout and anterograde amnesia, becomes common. - In a nontolerant person, a blood alcohol level over 400 mg/dL can produce respiratory failure, coma, and death. Due to tolerance,chronic heavy drinkers can present with fewer symptoms even with blood alcohol levels greater than 500 mg/dL.

Correct Answer. d

(53). A factory worker is required to submit to random drug tests as part of the “drug free policy” his employers have adopted. If he used cocaine five days before the test was administered, which assay is most likely to detect cocaine metabolites?

a. Semen

b. Hair

c. Saliva

d. Urine

Solution. (d) Urine. Ref– Read the text below Sol: - All the listed body fluids are used to detect cocaine use, except for semen. - Blood and saliva will provide the best level of current usage, while urine assay will detect use over the preceding several days. - Hair analysis can reveal the drug over weeks or months, but has little clinical applicability.

Correct Answer. d

(54). A man is convinced that his penis is receding into his body and fears that, when his penis disappears completely, he will die.Man is suffering from which culturebound Syndrome

a. Koro

b. Sangue dormido

c. Amok

d. Dha

Solution. (a) Koro Ref– Read the text below Sol: - Koro is seen in South and East Asia and is characterized by a sudden and intense anxiety connected to the belief that the penis is receding into the body and that death will follow when it has totally disappeared.

Correct Answer. a

a. 5-6 month of intrauterine life

b. 5-6 month of extra uterine life

c. 2nd year of life.

d. 5-6 years of life.

Solution. (c) 2nd year of life. Ref: Read the text below Sol: - At birth the mandible is in two separate halves united in the median plane by fibrous tissue. This union is known as symphysis menti. - After birth, in the first year, the two halves of the bone begin to join from below upwards. - A trace of separation may still be visible at the beginning of the second year. By the completion of two years, the two halves are completely joined.

Correct Answer. c

(56). Coronal suture is completely obliterated by

a. 25 years

b. 40 years

c. 55 years

d. 75 years

Solution. (b) 40 years Ref: Read the text below Sol: - Interestingly skull sutures obliterate earlier in males than in females. - Generally all bones ossify earlier in females, so this is an important exception. Suture closure begins at the inner table and progresses outwards. Generally there is a gap of 5 years before outer table also obliterates. - Coronal suture is completely obliterated by 40 years

Correct Answer. b

(57). Normal plasma osmolality is :

a. 275-300 milliosmol/kg

b. 310-340 millismol/kg

c. 200-240 milliosmol/kg

d. 160-180 milliosmol/kg

Solution. (a) 275-300 milliosmol/kg Ref.: Read the text below Sol : Plasma osmolality is a measure of the concentration of substances such as Nacl, K+, urea, glucose and other ions in the human plasma.

- It is calculated as osmules of solute per kg of solvent. - The normal plasma osmolality ranges between 280 to 303 milliosmoles per kg. - To calculate the plasma osmolality , the following formula can be used : Posm = 2[Na+]+[Glucose]/18+[BUN]/2.8

Correct Answer. a

a. Is part of the core enzyme.

b. Binds the antibiortic rifampicin.

c. Is inhibited by α-amanitin

d. Specifically recognizes promoter sites.

Solution. (d) Specifically recognizes promoter sites Ref– Read the text below Sol: - Sigma factor is required for correct initiation and dissociates from the core enzyme after the first bonds have been formed. - Core enzyme can transcribe but cannot correctly initiate transcription. - Rifampicin binds to the β subunit, and -amanitin is an inhibitor of eukaryotic polymerases.

Correct Answer. d

(59). Which of the following is the function of the large subunit of the ribosome?

a. Bind messenger RNA (mRNA)

b. Bind transfer RNA (tRNA)

c. Catalyze peptide bond formation

d. Link adjacent ribosomes in a polyribosome

Solution. (c) Catalyze peptide bond formation Ref– Read the text below Sol: - The large subunit of the ribosome catalyzes peptide bond formation by activation of peptidyl transferase. The small ribosomal subunit

contains the peptidyl-tRNA-binding (P) site that binds the tRNA molecule attached to the carboxyl end of the growing end of the polypeptide chain. - The small subunit also contains the aminoacyl-tRNA-binding (A) site that holds the incoming tRNA and amino acid. The initiation factors are loaded on the small ribosomal subunit that must locate the AUG (start) codon to initiate protein synthesis.

Correct Answer. c

(60). The drug, doxorubicin, is useful in the treatment of lymphomas and breast cancers because of its ability to interfere with which of the following enzyme activities?

a. DNA ligase

b. DNA polymerase-alpha

c. Primase

d. Topoisomerase II

Solution. -NA-

Correct Answer. d

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 26/86 (61). When fatty acids with odd numbers of carbon atoms are oxidized in the beta-oxidation pathway the final product is 1 mole of acetyl- CoA and 1 mole of the 3-carbon molecule,propionyl-CoA. In order to use the propionyl carbons, the molecule is carboxylated and converted ultimately to succinyl-CoA and fed into the TCA cycle. Which of the following represents the vitamin cofactor required in one of the steps of this conversion?

a. Cobalamin (B12)

b. Pantothenic acid (B5)

c. Pyridoxine (B6)

d. Riboflavin (B2)

Solution. (a) Cobalamin (B12) Ref– Read the text below Sol: - Propionyl-CoA is converted to succinyl- CoAin a series of reactions using three different enzymes.

- It is first carboxylated in an ATPdependent reaction catalyzed by propionyl- CoA carboxylase, an enzyme that requires biotin as a cofactor. - The product of the first reaction,d-methylmalonyl-CoA is then converted to l-methylmalonyl-CoA by methylmalonyl-CoA racemase. - Finally, methylmalonyl-CoA is converted to succinyl-CoA by the cobalaminrequiring enzyme, methylmalonyl-CoA mutase.

Correct Answer. a

(62). Which of the following apoproteins is found exclusively associated with chylomicrons?

a. Apo A

b. Apo B48

c. Apo CII

d. Apo D

Solution. (b) Apo B48 Ref– Read the text below Sol: - Apo B48 is found exclusively associated with chylomicrons and no other lipoprotein particle. Apo B48 is synthesized from an mRNA that is transcribed from the apo B100 gene. - Following transcription, the mRNA is edited within the intestinal epithelium yielding the apo B48 transcript. Apo(a) is an apoprotein found disulfide bonded to apo B100. - This then forms a complex with LDL, generating a novel lipoprotein particle identified as lipoprotein(a), Lp(a). Lp(a) has a strong resemblance to plasminogen and its presence in the circulation is highly correlated with premature coronary artery disease. - Apo CII is present in chylomicrons, VLDLs, LDLs, IDLs, and HDLs and is necessary for the activation of endothelial cell LPL. Apo D is found exclusively with HDLs and is also associated with cholesterol ester transfer protein (CETP) activity. - Apo E is found in chylomicrons,VLDLs, LDLs, IDLs, and HDLs. It is necessary for interaction of lipoprotein with the LDL-receptor (which is also referred to as the apo B100/apo E receptor).

Correct Answer. b

a. Acetyl-CoA

b. ATP

c. Dephosphorylation

d. NADH

Solution. (c) Dephosphorylation Ref– Read the text below Sol: - Regulation of the PDH complex is effected by both allosteric means and by phosphorylation . - Allosteric effectors of the enzyme complex are of both positive and negative type. The complex is phosphorylated by a specific kinase identified as PDH kinase and phosphate is removed by PDH phosphatase. When in the unphosphorylated state the complexis much more active, therefore the activity of PDH phosphatase is important for maintaining the PDH complex in the active state. - Acetyl-CoA and NADH are both allosteric inhibitors of the nonphosphorylated form of the PDH complex and also serves to activate PDH kinase leading to phosphorylated and inhibited PDH. - Like acetyl- CoA and NADH, ATP allosterically activates the PDH kinase leading to phosphorylated and inhbited PDH. Phosphorylation inhibits the activity of the PDH complex.

Correct Answer. c

(64). All of the following are correct about a primary transcript in eukartotes except that it :

a. Is usually longer than the functional RNA.

b. May contain nucleotide sequences that are not present in functional RNA.

c. Could contain information for more than one RNA molecule.

d. Contains a TATA box.

Solution. (d) Contains a TATA box Ref– Read the text below

Sol: - TATA box is part of the promoter, which is not transcribed.

Correct Answer. d

a. A frameshift mutation leading to premature termination of protein synthesis.

b. Mutation in the promoter region of the β- globin gene.

c. Mutation toward the 3’-end of the β-globin gene that codes for the polyadenylation site.

d. Mutation in the middle of an intron that is not at an a base.

Solution. (d) Mutation in the middle of an intron that is not at an a base Ref– Read the text below Sol: a) Mutation at either terminus of the intron would lead to a splicing error but a mutation in the middle of the intron should not. b) The exception would be if the specific mutation site were at the adenosine that forms the branch point.

Correct Answer. d

(66). Biological membranes are associated with all of the following except:

a. Prevention of free diffusion of ionic solutes.

b. Release of proteins when damaged.

c. Specific systems for the transport of uncharged molecules.

d. Free movement of proteins and nucleic acids cross the membrane.

Solution. (d) Free movement of proteins and nucleic acids cross the membrane. Ref– Read the text below Sol: - These molecules are too large to cross the membrane freely unless the membrane is damaged

Correct Answer. d

(67). Basal lamina structure is produced by connecting planar networks of laminins and type IV collagen. Which of the following statements is untrue correct?

a. Both laminin and type IV collagen are composed of four polypeptide chains.

b. Laminin has a cruciform structure.

c. Laminin and type IV collagen are interconnected by a heparinsulfate proteoglycan.

d. Type IV collagen contains repeating sequences of ( Gly-HyProY).

Solution. (a) Both laminin and type IV collagen are composed of four polypeptide chains. Ref– Read the text below

Sol: - Both laminin and type IV collagen are composed of three polypeptide chains

Correct Answer. a

a. Chemical identity of the substrate.

b. The atomic mass of the elements in the reactive

c. Optical activity of product formed from a symmetrical substrate.

d. Type of reaction catalyzed.

Solution. (b) The atomic mass of the elements in the reactive group Ref– Read the text below Sol:

- Enzymes do not distinguish among different nuclides of an element,although the rate of reaction of a heavier nuclide might be less than that of a lighter one.

Correct Answer. b

(69). At which of the following enzyme-catalyzed steps of the tricarboxylic acid cycle does net incorporation of the elements of water into an intermediate of the cycle occur ?

a. Aconitase

b. Citrate synthase

c. Malate dehydrogenase

d. None

Solution. (b) Citrate synthase Ref– Read the text below Sol:

- Water is required to hydrolyze the thioester bond of acetyl CoA. - Aconitase removes water and then adds it back. - The dehydrogenases remove two protons and two elecrrons.

Correct Answer. b

(70). All of the following tricarboxylic acid cycle intermediates may be added or remove by other metabolic pathways except:

a. Citrate.

b. Fumarate.

c. Isocitrate.

d. D α-ketoglutarate.

Solution. (c) Isocitrate. Ref– Read the text below Sol: - Citrate is transported out of the mitochondria to be used as a source of cytoplasmic acetyl CoA. - Fumarate is produced during phenylalanine and tyrosine degradation.

Correct Answer. c

a. NADH.

b. Acetyl CoA.

c. GTP.

d. ATP.

Solution. (d) ATP. Ref– Read the text below

Sol: - ATP and ADP are transported in opposite directions.

Correct Answer. d

(72). All of the following events are usually involved in the synthesis of triacylglycerols in adipose tissue except:

a. Addition of a fatty acyl CoA to diacylglycerol.

b. Addition of a fatty acyl CoA to a lysophosphatide.

c. A reaction catalyzed by glycerol kinase.

d. Hydrolysis of phosphatidic kinase

Solution. (c) A reaction catalyzed by glycerol kinase. Ref– Read the text below Sol:

- Option C does not occur to any significant extent in adipose tissue. - The sequential addition of fatty acyl CoAs to glycerol 3-phosphate forms lysophosphatidic acid and then forms phosphatidic acid, whose phosphate is removed before the addition of the third fatty acyl residue.

Correct Answer. c

(73). Thiamine pyrophosphate is coenzyme required for :

a. Branched – chain amino acid dehydrogenase complex

b. 2-hydroxyphytanoyl-CoA lyase

c. Transketolase

d. All the above

Solution. (d) All the above Ref.: Read the text below Sol : Thiamine pyrophosphate or thiamine diphosphate (ThDP) is a thiamine derivative which is cleaved by thiamine pyrophosphatase. ThDP is a coenzyme to many enzymes, such as :

- Pyruvate dehydrogenase complex. - Alpha-ketoglutarate dehydrogenase complex. - Branched – chain amino acid dehydrogenase complex. - 2-hydroxyphytanoyl-COA lyase. - Transketolase.

Correct Answer. d

a. Abdomen

b. Thorax

c. Hand

d. Forehead

Solution. (d) Forehead Ref: Read the text below Sol: - Whereever in the body, the skin lies directly over the bone, with a scanty layer of fat in between, lacerated wounds appear like incised wounds. - These are known as incised looking lacerated wounds. - Other areas where lacerated wounds would appear like an incised wound are scalp, chin, eyebrows, lower jaw, iliac crest and shin.

Correct Answer. d

(75). Which of the following tissue suffers most in a blast caused by an explosion?

a. Lung

b. Liver

c. Skeletal system

d. Nervous system including brain

Solution. (a) Lung Ref: Read the text below Sol: - Blast caused by an explosion is a wave of compression followed by a transient zone of low pressure (below atmospheric pressure). So a rapid double change in pressure is suffered by the body. A blast causes the most damage at an interface between tissues in contact with

the atmosphere. This is why lung suffers the most. - Other internal organs having an interface with air are (ii) middle ear and (iii)gastrointestinal tract. These two suffer greatly in explosions.

Correct Answer. a

(76). Which of the following tissues is most resistant to electric current?

a. Muscle

b. Skin

c. Bone

d. Blood

Solution. (b) Skin Ref: Read the text below Sol: - The most resistant tissular layer of the body is the skin, followed in order of decreased resistance, by bone, fat, nerve, muscle, blood, and body fluids. - Variations in resistance are primarily determined by the water content of the tissues. - Dry skin has a higher resistance than skin moist with sweat.

Correct Answer. b

a. Liver

b. Spleen

c. Lungs

d. Brain

Solution. (d) Brain Ref: Read the text below Sol: - Blunt force that causes sudden angular deceleration or acceleration of the brain can often cause Diffuse Axonal Injury (DAI). - The axons are broken; the broken axons retract into a ball like structure, known as “retraction balls” (or “axonal spheroids”). - The presence of “Retraction balls” thus indicate axonal damage. They are seen microscopically in the white matter usually after a few days of injury.

Correct Answer. d

(78). “Undertaker’s fracture” is commonly seen in:

a. Skull

b. Cervical spine

c. Lumbar spine

d. Pelvis

Solution. (b) Cervical spine Ref: Read the text below Sol: - Undertaker’s fracture is basically a postmortem fracture, and occurs due to careless handling of the dead body by undertakers; in fact

this fracture can be produced by anyone who has to do anything with the dead body. - Pathologists are known to extend the head to make the removal of the neck structures easier. To do this, they usually insert a block about 10-15 cm high under the neck may fall forcibly backwards, producing this fracture. - It involves subluxation of the lower cervical spine due to tearing of the intervertebral disc at about C6-C7

Correct Answer. b

(79). After postmortem examination, the body has to be handed over to the

a. Investigating officer

b. The authorities of the nearest crematorium or burial ground

c. Medical superintendent of the hospital in which the post mortem was conducted.

d. Relatives

Solution. (a) Investigating officer Ref: Read the text below Sol: - It is important to remember that the doctor hands over the body to the investigating officer only (to the police in cases of police inquest and to the magistrate in cases of magisterial inquest). - It is the investigating officer who passes on the body to the relatives. - The relatives then take the body to the crematorium or burial ground.

Correct Answer. a

a. Remnants of genital organs help to determine sex because they resist putrefaction

b. Peeled off skin cannot be used for fingerprinting

c. Hair, its colour and dyeing when present can be used for identification

d. Body and dental X-rays are useful in identification.

Solution. (b) Peeled off skin cannot be used for fingerprinting Ref: Read the text below

Sol: - Peeled off skin can be used for finger printing.

Correct Answer. b

(81). Winslow’s test was once used to detect the stoppage of :

a. Brain function

b. Liver function

c. Respiration

d. Circulation.

Solution. (c) Respiration Ref: Read the text below Sol: - This test was suggested by the Danish physician Jacques – Benigne Winslow. It consisted of keeping a small pot containing water or mercury over the thoraco-abdominal region (usually just at or below xiphisternum). - A ray of light was allowed to fall over the liquid. If the respiration was taking place, the ray of light (over the surface of the liquid) would appear to move. - The three tests – magnus, icard and winslow – however weird they may appear gained prominence because the stethoscope had not been developed fully and in cases of suspended animation, one could easily mistake a living person for dead.

Correct Answer. c

(82). If a person is not heard of for a certain number of years by those who would naturally have heard of him, he is supposed to have been dead, and the burden of proving that he is alive shifts on the person who affirms it. For how many years, he should not have been heard of?

a. 4 years

b. 7 years

c. 15 years

d. 30 years

Solution. (b) 7 years Ref: Read the text below Sol: - This is according to section 108 of the IEA. This is known as presumption of Death. Here is the full statement of Sec 108 of IEA S 108. - Burden of proving that person is alive who has not been heard of for seven years.

Correct Answer. b

a. It is a direct precursor of guanine

b. It covalently binds to allopurinol

c. It is oxidized to form uric acid

d. It is oxidized to form hypoxanthine

Solution. (c) It is oxidized to form uric acid Ref– Read the text below Sol:

- Xanthine oxidase catalyzes the last two steps in the degradation of purines. - Hypoxanthine is oxidized to xanthine, and xanthine is further oxidized to uric acid. Thus, xanthine is both product and substrate in this two-step reaction. In humans, uric acid is excreted via the urine.

Correct Answer. c

(84). Which of the following statements about solutions of amino acids at physiologic pH is true?

a. All amino acids contain both positive and negative charges

b. All amino acids contain positively charged side chains

c. Some amino acids contain only positive charges

d. All amino acids contain negatively charged side chains

Solution. (a) All amino acids contain both positive and negative charges Ref– Read the text below Sol: - At neutral pH, amino acids in solution are zwitterions (i.e., dipolar ions) containing both a protonated amino group (pK approximately 9.5) and a dissociated carboxyl group (pK approximately 2).

- At pH 7.4, the pH −pK from the Henderson-Hasselbalch equation is ~5 for the carboxy group, predicting a ratio of base (carboxyl anion) to acid (carboxylic acid) of 105. - Similarly, the pH −pK for the amino group is about −2, predicting a ratio of base (amino group) to acid (protonated ammonium ion) of less than 102. - Amino acids with ionizable side chains may have charges in addition to those of the amino and carboxyl groups.

Correct Answer. a

(85). Immunoglobulin G molecules can be characterized by which of the following statements?

a. They are maintained at a constant level in the serum

b. They contain nucleic acids

c. They contain mostly carbohydrate

d. They can be separated into subunits with a reducing agent and urea

Solution. (d) They can be separated into subunits with a reducing agent and urea Ref– Read the text below Sol: -- Immunoglobulin G is composed of pairs of light chains and heavy chains attached by disulfide bridges. - If the reducing agent mercaptoethanol is used to break the disulfide bridges and urea is used to disrupt noncovalent interactions, two identical light subunits (25 kd) and two identical heavy chains (50 kd) per protein can be resolved with electrophoresis.A small amount of carbohydrate is also present. - In contrast,the proteolytic enzyme papain cleaves the heavy chains, which results in two Fab molecules consisting of the entire light chain attached to the amino terminal half of each heavy chain and two Fc molecules consisting of the carboxyl terminal half of each heavy chain. - Other proteolytic enzymes are nonspecific. Levels rise and fall in the serum dependent upon specific induction by antigen.

Correct Answer. d

a. Dialysis

b. Affinity chromatography

c. Gel filtration chromatography

d. Ion exchange chromatography

Solution. (b) Affinity chromatography Ref– Read the text below Sol: - Each of the techniques listed separates proteins from each other and from other biologic molecules based upon characteristics such as size, solubility, and charge. - However, only affinity chromatography can use the high affinity of proteins for specific chemical groups or the specificity of immobilized antibodies for unique proteins. - In affinity chromatography,a specific compound that binds to the desired protein—such as an antibody, a polypeptide receptor, or a

substrate—is covalently bound to the column material. - A mixture of proteins is added to the column under conditions ideal for binding the protein desired, and the column is then washed with buffer to remove unbound proteins. - The protein is eluted either by adding a high concentration of the original binding material or by making the conditions unfavorable for binding (e.g., changing the pH).The other techniques are less specific than affinity binding for isolating proteins. - Dialysis separates large proteins from small molecules. Ion exchange chromatography separates proteins with an overall charge of one sort from proteins with an opposite charge (e.g., negative from positive).Gel filtration chromatography separates on the basis of size. - Electrophoresis separates proteins on the principle that net charge influences the rate of migration in an electric field.

Correct Answer. b

(87). Which one of the following proteins is found in the thick filaments of skeletal muscle?

a. α-actinin

b. Myosin

c. Troponin

d. Tropomyosin

Solution. (b) Myosin Ref– Read the text below Sol: - Two kinds of interacting protein filaments are found in skeletal muscle. - Thick filaments 15 nm in diameter contain primarily myosin. Thin filaments 7 nm in diameter are composed of actin, troponin, and

tropomyosin. - The thick and thin filaments slide past one another during muscle contraction. - Myosin is an ATPase that binds to thin filaments during contraction. α-actinin can be found in the Z line.

Correct Answer. b

a. Trypsin

b. Chymotrypsin

c. Elastase

d. Pepsin

Solution. (d) Pepsin Ref– Read the text below Sol: - Pepsin is secreted in a proenzyme form in the stomach. - Unlike the majority of proenzymes, it is not activated by protease hydrolysis. - Instead, spontaneous acid hydrolysis at pH 2 or lower converts pepsinogen to pepsin. Hydrochloric acid secreted by the stomach lining creates the acid environment. All the enzymes secreted by the pancreas are activated at the same time upon entrance into the duodenum. - This is accomplished by trypsin hydrolysis of the inactive proenzymes trypsinogen, chymotrypsinogen, procarboxypeptidase, and proelastase. - Primer amounts of trypsin are derived from trypsinogen by the action of enteropeptidase secreted by the cells of the duodenum.

Correct Answer. d

(89). Thiamine deficiency is due to

a. Alcoholism

b. Prolonged breast feeding

c. Chronic diuretic

d. All of these

Solution. (d) All of these Ref:Read the text below Sol: - Vitamin B1 deficiency is usually connected to alcoholism, malabsorption diseases, and poor diet. - Thiamine is a coenzyme for the decarboxylation of pyruvate and the oxidation of alpha keto-glutamic acid. Lipoic acid which is formed in the liver is also required for the reactions. Patients with liver disease may show signs of B1 deficiency.

- The minimum amount of thiamin needed by the body can increase because of increased physiological or metabolic demands arising from pregnancy and lactation, heavy physical exertion, illnesses like cancer, liver diseases, hyperthyroidism, and surgery. - Thiamin deficiency occurs as a result of many factors, including crash dieting, alcohol abuse, liver disfunction, kidney dialysis, and sustained periods of IV nutrients. - Also at risk are those who consume a lot of sweets, soft drinks, and highly processed foods. Vitamin B1 deficiency is common among alcoholics, as chronic alcohol consumption decreases the amount of Vitamin B1 absorbed by the body

Correct Answer. d

a. Cortisol

b. ACTH

c. ANP

d. Adrenaline

Solution. (d) Adrenaline Ref:Read the text below Sol: - Several important peptide hormones are secreted from the pituitary gland. - The anterior pituitary secretes prolactin, which acts on themammary gland, adrenocorticotrophic hormone (ACTH), which acts on the

adrenal cortex to regulate the secretion of glucocorticoids, andgrowth hormone, which acts on bone, muscle, and the liver. - The posterior pituitary gland secretes antidiuretic hormone, also called vasopressin, and oxytocin. - Peptide hormones are produced by many different organs and tissues, however, including the heart (atrial-natriuretic peptide (ANP) or atrial natriuretic factor (ANF)) and pancreas (insulin and somatostatin), the gastrointestinal tract cholecystokinin,gastrin), and adipose tissue stores (leptin)

Correct Answer. d

(91). Addis test is used for –

a. Urinary sugar

b. Urinary sediment

c. Blood lactate

d. Serum magnesium

Solution. (b) Urinary sediment Ref:Read the text below Sol: Addis count: - The Addis count is a quantitative measurement of red blood cells, white blood cells, and casts in a 12-hour overnight urine specimen. - Protein and specific gravity may also be included. Fluids may be restricted before the test so that urine will be concentrated. - The Addis test is used to evaluate the course of renal disease by comparing results over time.

Correct Answer. b

(92). The minimum amount of air usually accepted to be able to cause fatal air embolism in adults is

a. 5ml

b. 100ml

c. 250ml

d. 500ml

Solution. (b) 100ml Ref: Read the text below Sol:

- The volume of air needed to cause fatal embolism has been hotly debated for years, but no real consensus has emerged. - The figures vary from 10 to 480 ml. if the volume of the right side of the heart is accepted as the minimum space that has to be filled, about 100ml would appear to be reasonable volume.

Correct Answer. b

a. Cranial cavity

b. Chest cavity

c. Abdominal cavity

d. Uterus

Solution. (a) Cranial cavity Ref: Read the text below Sol: - Heat hematoma is formed in the Extradural space. - The blood in heat haematoma comes either from venous sinuses - The blood from diploe, supposedly “boils out” due to great heat and passes through emissary veins, to come to lie just beneath the skull (and outside the dura.)

Correct Answer. a

(94). How many points of similarity between two fingerprints have to be established before the two can be said to be identical?

a. 6

b. 16

c. 25

d. The figure varies from country to country.

Solution. (d) The figure varies from country to country. Ref:Read the text below Sol: - Fingerprints may be deposited in natural secretions from the eccrine glands present in friction ridge skin (secretions consisting primarily of water) or they may be made by ink or other contaminants transferred from the peaks of friction skin ridges to a relatively smooth surface such as a fingerprint card. - The term fingerprint normally refers to impressions transferred from the pad on the last joint of fingers and thumbs, though fingerprint cards also typically record portions of lower joint areas of the fingers (which are also used to make identifications). - Number of points of similarity between two fingerprints have to be established before the two can be said to be identical always varies from country to country.

Correct Answer. d

(95). Mental ability of a person to make a valid will is known as :-

a. Compos Mentis

b. Testamentary capacity

c. Corpus delicti

d. Corona

Solution. (b) Testamentary capacity Ref:Read the text below

Sol: - Will denotes any testamentary document (S.31, I.P.C.)

Correct Answer. b

a. Immediately after death, the body may shorten by about 2-3 cm.

b. Immediately after death, the body may lengthen by about 2-3cm.

c. Change is stature after death occurs only when the deceased was engaged in struggle at the time of death.

d. There is absolutely no change in stature after death.

Solution. (b) Immediately after death, the body may lengthen by about 2-3cm. Ref:Read the text below

Sol: - Immediately after death, the body may lengthen by about 2-3cm.

Correct Answer. b

(97). A person is called a Juvenile or child when he is below the age of :

a. 12 years

b. 14 years

c. 16 years

d. 17 years

Solution. (d) 17 years Ref:Read the text below Sol: - The Juveniles Act defines as a "juvenile" a person under the age of 17 years and subdivides this group into

"child", meaning a person under the age of 14 years, and "young person", meaning a person who has attained the age of 14 years and is under the age of 17 years. - This Act provides for the minimum age at which a person has criminal liability, matters relating to the employment of children, giving testimony in court, and the consumption of alcohol by children.

Correct Answer. d

(98). The age of criminal liability is

a. 12 years

b. 14 years

c. 16 years

d. 17 years

Solution. (a) 12 years Ref:Read the text below Sol:

- The age of criminal liability is 12 years. - In the words of section 3 of the Juveniles Act, "It shall be conclusively presumed that no child under the age of 12 years can be guilty of any offence."

Correct Answer. a

a. Nicotinic acid

b. Enalapril

c. Meperidine

d. Chloroquine

Solution. (b) Enalapril Ref:Read the text below Sol: DRUGS CAUSING TOXIC MYOPATHIES Lipid-lowering agents Fibric acid derivatives ,HMG-CoA reductase inhibitors Niacin (nicotinic acid) Glucocorticoids

Nondepolarizing neuromuscular blocking agents Zidovudine Drugs of abuse Alcohol, Amphetamines, Cocaine Heroin, Phencyclidine, Meperidine Autoimmune toxic myopathy- D-Penicillamine Amphophilic cationic drugs Amiodarone, Chloroquine, Hydroxychloroquine Antimicrotubular drugs Colchicine

Correct Answer. a

(100). Choose the correct statement:

a. Rape is defined under section 3761PC

b. Section 228 Cr. PC prohibits disclosure of the identity of the victims of rape

c. Under sections 377 Cr. Pc trials of rape shall be conducted in camera

d. Section 374 IPC lays down the punishment for offence of Rape

Solution. (b) Section 228 Cr. PC prohibits disclosure of the identity of the victims of rape Ref:Read the text below Sol: RAPE Rape is generally declined as unlawful sexual intercourse by a man with any women against her wall,without her consent, or with her consent when it has been obtained by unlawful means LAW ON RAPE IN INDIA Under section 375 IPC, rape is defined as unlawful sexual intercourse by a man his wife under the age of 15

years, with any other women under the age of 16 years or above that age, against her will, without her consent, or with her consent when it has been obtained by unlawful means Exception: Sexually intercourse by a man with his wife, even against her will, is not rape, she is above 15 years of age. Section 376 IPC lays down the punishment for the offence of rape, which may extend from seven years to life imprisonment and also fine. Section 228 IPC prohibits disclosure of the identity of the victim of rape Under section 327 Cr. PC, the inquiry into and trial of rape or an offence under section 376 IPC shall be conducted in camera and it is not lawful for any person to print to or publish any matter in relation to such proceedings except with the permission of the Court

Correct Answer. b

(101). What causes post mortem luminescence?

a. Dhatura

b. Mercury

c. Armillaria

d. Oleander

Solution. (c) Armillaria Ref: Read the text below Sol:

Causes for post mortem luminescence 1. Armillaria millea/ Ramsbottom 2. Photobacterium fischerii

Correct Answer. c

(102). Acute arsenic poisoning viscera preserved is

a. Liver, stomach and blood

b. Liver, stomach, blood, kidneys, intestine, urine

c. Routine viscera and hairs

d. Routine viscera and hair, nails, bones and skin

Solution. (a) Liver, stomach and blood Ref: Read the text below

Sol: - In chronic arsenic poisoning we preserve routine viscera and hair, nails, bones and skin

Correct Answer. a

(103). Which one of the enzyme deficiency is correctly paired?

a. Fabry’s – alpha galactosidase

b. Gauchers – beta glucosidase

c. Cystic fibrosis – glucoronidase

d. Chronic bronchitis – alpha I antitrypsin

Solution. (a) Fabry’s – alpha galactosidase Ref.: Read the text below Sol : - This X-linked recessive condition is due to a deficiency of the lysosomal hydrolase α-galactosidase causing an accumulation of glycosphingolipids with terminal α–galactosyl moieties in the lysosomes of various tissues including the liver, kidney, blood vessels and the ganglion cells of the nervous system. - The patients present with peripheral nerve involvement, but eventually most patients develop renal problems in adult life.

Correct Answer. a

(104). In collagen synthesis, hydroxyproline is formed from:

a. Prolene

b. Lysine

c. Hydroxylysine

d. None of the above

Solution. (a) Prolene Ref.: Read the text below Sol :

- During collagen formation, proline is hydroxylated in position 4 (when situated at the amino-side of the glycine) or in position 3 (when situate at the carboxy-side of the glycine). Lysine is hydroxylated in position 5.

Correct Answer. a

a. Integrin is involved in fibronectin transport

b. Erythrocyte band III protein is involved in bicarbonate transport

c. Na+-K+- ATPase is involved in cation transport

d. Most membrane transport proteins are transmembrane protein

Solution. (a) Integrin is involved in fibronectin transport Ref.: Read the text below Sol : - Erythrocyte band III protein has a hydrophilic core for ion transport and is involved in bicarbonate transport. - Na+-K+-ATPase uses energy stored in ATP to maintain the surface potential differences between the cell’s inner and outer surfaces and is involved in cation transport. Most membrane proteins are amphipathic.

- Most membrane proteins are amphipathic. - Transmembrane proteins such as integrin span the entire membrane. They have hydrophilic domains associated with the water inside and outside the cell. - They also have a hydrophobic domain associated with the hydrophobic portions of phospholipids. - Integrin is a membrane receptor for extracellular matrix molecules such as fibronectin and laminin. It binds with extracellular fibronectin to anchor the cell to the extracellular matrix, but it does not transport fibronectin.

Correct Answer. a

(106). The transfer of genetic material from one bacteria to other through bacteriophage is known as

a. Conjugation

b. Transcription

c. Transduction

d. Translation

Solution. (c) Transduction Ref.: Read the text below Sol : - Occasionally, the virus carries a portion of host DNA

- Subsequent infection of a new host by such a virus may introduce new genes to the new host. This is called viral transduction. - This process was studied by Joshua Lederberg (Nobel Prize 1958) and Max Delbruck (Nobel Prize 1969. - This can be considered as genetic engineering by nature, or horizontal transmission of genes.

Correct Answer. c

(107). Abnormal base in t RNA is :

a. Dihydrouracil

b. Orotic acid

c. Methyl Xanthine

d. Cystine

Solution. (a) Dihydrouracil Ref.: Read the text below Sol : - The transfer RNAs show extensive internal base pairing and acquire clover leaf like structure.

- They contain a significant proportion of unusual bases. - These include dihydrouracil (DHU), pseudouridine and hypoxantine. - Many bases are methylated ; this occurs in the nucleus. - The 5’ end often has a phosphorylated guanosine.

Correct Answer. a

a. Increased Km and Vmax is same

b. Increased Vmax and Km remains constant

c. Increased Km and increased Vmax

d. Decreased Km and increased Vmax

Solution. (a) Increased Km and Vmax is same Ref.: Read the text below Sol : Competive inhibition

- The inhibitor competes with the substrate for binding at the enzymes active site. - Km therefore increases, since a higher concentration of inhibitor decreases substrate binding and more substrate is needed to reach ½ Vmax. - Vmax remains unchanged since the dissociation of the enzyme – substrate complex, once formed, is unaffected.

Correct Answer. a

(109). A genetic disorder renders fructose 1,6 bisphosphatase in the liver less sensitive to regulation by fructose 2,6 bisphosphate. All of the following metabolic changes are observed in this disorder except ?

a. Level of fructose 1, 6-bisphosphate is higher than normal

b. Level of fructose 1, 6-bisphosphate is lower than normal

c. Less pyruvate will be formed

d. Less ATP will be generated

Solution. (a) Level of fructose 1, 6-bisphosphate is higher than normal Ref.: Read the text below Sol : - Fructose 2, 6-bisphosphate has got inhibitory effect on fructose 1, 6-bisphosphatase enzyme. It inhibits fructose 1, 6-bisphosphatase by increasing the K m for fructose 1, 6-bisphosphate. - Fructose 2, 6-bisphosphate is formed by phosphorylation of fructose 6-phsphate by phosphofructokinase-2. - The same enzyme is also responsible for its breakdown, since it has fructose 2, 6-bisphosphatase activity. - This bifunctional enzyme is under the allosteric control of fructose 6-phosphate,which stimulates the kinase and inhibits the phosphatase. - Hence, when there is an abundant supply of glucose, the concentration of fructose 2, 6-bisphosphate increase, stimulating glycolysis by activating phosphor-fructokinase – 1 and inhibiting fructose 1, 6-bisphosphatase

Correct Answer. a

(110). Regarding conversion of pyruvate to acetyl Co A and CO2 , true statement is ?

a. Reversible reaction

b. Lipoic acid is required

c. Activated when PDH complex is phosphorylated

d. Cytosolic reaction

Solution. (b) Lipoic acid is required Ref.: Read the text below

Sol : - PDH reaction is irreversible reaction. It is active in dephosphorylated form. This reaction takes place within the mitochondria.

Correct Answer. b

a. Glyceraldehyde 3 phosphate dehydrogenase

b. Phosphoglucomutase

c. Lactate dehydrogenase

d. Glucokinase

Solution. (a) Glyceraldehyde 3 phosphate dehydrogenase Ref.: Read the text below Sol : - This enzyme is very important for converting glyceraldehydes 3 phosphate to 1, 3 bisphosphoglycerate during first step of oxidoreduction phase of glycolysis.

- Phosphoglucomutase is involved in glycogenesis and glycogenolysis for interconversion of glucose 6 phosphate and glucose 1 phosphate. - Lactate dehydrogenase is involved in interconversion of pyruvate and lactate. - Glucokinase is needed to metabolise glucose. - This enzyme is not needed for metabolizing fructose.

Correct Answer. a

(112). After overnight fast, level of glucose transporter are reduced in?

a. Brain cell

b. Hepatocyte

c. Adipocyte

d. RBCs

Solution. (c) Adipocyte Ref.: Read the text below Sol : - Insulin is needed for GLUT mediated uptake of glucose in skeletal muscle, cardiac muscle and adipose tissue via GLUT-4. - So in these organs the expression of GLUT will be affected after an overnight fast. - Brain cells, liver cells and RBC do not require insulin for glucose transportation.

Correct Answer. c

(113). Acetyl Co A can be converted into all of the following except ?

a. Glucose

b. Fatty acid

c. Cholesterol

d. Ketone bodies

Solution. (a) Glucose Ref.: Read the text below Sol : - Acetyl Co A cannot enter gluconeogenic pathway as PDH reaction is irreversible and there is no reaction which can for pyruvate from acetyl Co A.

- Acetyl Co A is the precursor molecule for fatty acid synthesis acetyl Co A is converted to malonyl Co A by rate limiting enzyme acetyl Co A carboxylase, which then enters in the cyclical reaction of fatty acid synthesis as a precursor molecule. - Acetyl Co A is the precursor molecule for choleasterol synthesis. - Ketone body synthesis require acetyl Co A from fatty acid oxidation.

Correct Answer. a

a. HDL

b. LDL

c. Chylomicron remnant

d. VLDL

Solution. (a) HDL Ref.: Read the text below Sol : Distribution of various apolipoprotein on lipoprotein is as follow : - Chylomicrons A-I, A-II, A-IV, B-48, C-I, C-II, C-III, E - Chylomicron remnants B – 48, E - VLDL B-100, C-I, C-II, C-III, E - IDL B-1000, E - LDL B-100 - HDL A-I, A-II, A-IV, C-I, C-II, C-III, D, E - Pre -β- HDL A-I

Correct Answer. a

(115). Plant penicillin is

a. DDT

b. Endrin

c. Lindane

d. Ergot

Solution. (b) Endrin Ref: Read the text below Sol: - Endrin a.k.a Plant penicillin is an organochlorine. It is a polycyclic, polychlorinated hydrocarbon.

- It is called as Plant penicillin because of its broad spectrum of activity against various insect pests. - Features of poisoning are similar to organophosphorous compound poisoning. - On post mortem examination smell of kerosene may be appreciated. - Endrin resists putrefaction and can be detected in the viscera quite some time after death.

Correct Answer. b

(116). Hippus is seen in

a. Aconite

b. Alcohol

c. Diazepam

d. Strychnine

Solution. (a) Aconite Ref: Read the text below Sol: Aconite (mitha zaher, monk’s hood) – Cardiac poison All the parts of the plant are poisonous. Fatal dose is 1g of dried root

C/F – 1. Pain and tingling all over the body 2. headache, giddiness, pallor, profuse sweating, subnormal temperature. 3. paraesthesia, ataxia and slurring of speech 4. cardiac arrhythmias, AV Block 5. Hippus – alternate contraction and dilatation of the pupil.

Correct Answer. a

a. Dead born in 12 hours

b. Live born in 12 hours

c. Dead born in 24 hours

d. Live born in 24 hours

Solution. (a) Dead born in 12 hours Ref: Read the text below Sol: Mummification: Aseptic autolysis Robert’s sign: presence of gas in the great vessels, seen in 12 hours of intrauterine death Spaulding sign: overriding of bones of cranial vault, seen in 5 -7 days of intrauterine death.

Correct Answer. a

(118). The Munchausen syndrome by proxy was given by

a. Robert Haughen

b. Rosenberg

c. Locard

d. Burke

Solution. (b) Rosenberg Ref: Read the text below Sol: Rosenberg gave four diagnostic criteria for Munchausen syndrome by proxy

1. Illness produced or alleged, or both by a parent 2. Repeated requests for medical care of a child, leading to multiple medical procedures 3. Parental denial of knowledge of the cause of symptoms 4. Regression of symptoms when the child is separated from the parents

Correct Answer. b

(119). Prescription drugs are given in which schedule in Drugs and Cosmetics Rules 1945.

a. C

b. F

c. H

d. X

Solution. (c) H Ref: Read the text below Sol: Drugs and Cosmetics Rules 1945 framed under the Drugs act 1940 have classified the drugs in various schedules as follows c- biological and special products F- vaccines and sera H - prescription drugs X – potential drugs of abuse

Correct Answer. c

a. Ciguatera poisoning

b. Scombroid poisoning

c. Venomous fish

d. Box jelly fish

Solution. (b) Scombroid poisoning Ref: Read the text below Sol: Scombroid food poisoning is a foodborne illness that results from eating spoiled fish. - So called because it was first noted with fishes of the suborder Scombroidea (mackerels, tunas, albacores, bonitos, swordfishes, sailfishes). However it has been noted with nonscombroid fish too (mahi-mahi and amberjack). - Second most common type of seafood poisoning, second only to ciguatera poisoning.

- Cause - Histidine present in several fishes (especially in those having dark meat). At > 16°C it is converted to histamine via histidine decarboxylase produced by Morganella morganii (a Gram-negative bacillus). This is why fish should be stored at low temperatures. Histamine is not destroyed by normal cooking temperatures, so even properly cooked (spoiled) fish are poisonous. - Scombroid syndrome - Histamine produces allergy like reactions eg. skin flushing, throbbing headache, oral burning, abdominal cramps, palpitations (Scombroid syndrome). Symptoms usually occur within 10-30 minutes of ingesting the fish and generally are self-limited. D/D - allergic reactions

Correct Answer. b

(121). Privileged communications means exception to the general rule of:

a. Res ipsa loquitur

b. Novus actus intraveniens

c. Professional Secrecy

d. Professional Negligence

Solution. (c) Professional Secrecy Ref:Read the text below Sol: PRIVILEGED COMMUNICATION - It is a statement made bonafide upon any subject matter by a doctor to the authority, due to his duty to protect the interests of the community or of the State. - Privileged communication are the exceptions to the general rule of professional secrecy. - Professional secrecy, the first concept in the Code of Professional Ethics, is intended to protect the conference organiser against any leak of confidential information both during and after the meeting.

Correct Answer. c

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 48/86 (122). If a mentally ill person is detained against the provision of the mental Act, the responsible can be punished with :

a. Imprisonment to 12 months

b. Imprisonment upto 6 months

c. Imprisonment upto 24 months

d. No imprisonment

Solution. (c) Imprisonment upto 24 months Ref:Read the text below Sol: THE MENTAL HEALTH ACT, 1987 Main provisions are: - (“Mentally ill persons” is defined as” a person who is in need of treatment by reason of any mental disorder other than mental retardation. - The Central Government and State Governments have to establish an Authority for Mental Health for regulation,development, direction and coordination with respect to Mental Health Services. - Establish and maintenance of psychiatric hospitals or psychiatric nursing homes can be done only with license,which has to be renewed every five year. - If a mentally ill person is received or detained against the provisions of the Act. the punishment is imprisonment up to two years and also fine. - Not less than three visits shall make a joint inspection of hospital or nursing home at least once in every minor parent.

Correct Answer. c

(123). A person who may be impotent with one particular woman is called :-

a. Quoad

b. Frigidity

c. Atavism

d. None of the above

Solution. (a) Quoad Ref:Read the text below

Sol: A person who may be impotent with one particular woman is called quoad.

Correct Answer. a

(124). Dribbling of saliva is seen in a case of

a. Hanging

b. Strangulation

c. Drowning

d. Suffocation

Solution. (a) Hanging Ref.: Narayan Reddy’s-160. Sol : - In hanging, saliva may be found dribbling from the angle of the mouth when the head is drooping forward. This is due to increased salivation just before death due to the stimulation of the salivary glands by the ligature. - Slight hemorrhage or bloody froth is sometimes seen at the mouth or nostrils.

Other signs of hanging are - Ligature mark around the neck. - Presence of ecchymoses, abrasions and redness around the ligature mark - Ecchymoses of the larynx or epiglottis. - Rupture of the intima of the carotid - Postmortem signs of asphyxia.

Correct Answer. a

a. Before 6 weeks

b. Before 16 weeks

c. Before 26 weeks

d. Before full term

Solution. (d) Before full term Ref.: Narayan Reddy’s-200. Sol : - Legally, abortion (miscarriage) is the premature expulsion of the fetus from the mother’s womb at any time of pregnancy before the term of pregnancy is completed.

Correct Answer. d

a. Single edged knife

b. Double edged knife

c. Bayonet

d. None

Solution. (a) Single edged knife Ref: Read the text below Sol: - Stab by a single edged weapon causes fish tailing wound - Stab by a double edged weapon causes spindle shaped wound.

Correct Answer. a

a. Dhatura

b. Opium

c. Arsenic

d. Sulfuric acid

Solution. (a) Dhatura Ref.: Robbin’s - 686-87 Sol : - The seeds and fruit of dhatura contain traces of atropine which leads to dilated pupil. Dilated pupils are seen in poisoning by – - Amphetamine - Antihistamines - Botulin toxin - Carbon monoxide - Cocaine - Cyanide - Dhatura - Ephedrine

Correct Answer. a

(128). The cause of death in hanging is

a. Asphyxia

b. Cerebral anoxia

c. Fracture/dislocation of cervical vertebrae

d. All of the above

Solution. (d) All of the above Ref.: Narayan Reddy’s, -157. Sol : - Causes of death in hanging may be due to : - Asphyxia : blockage of the airway (cerebral anoxia) by the constricting force of the ligature (a tension of 15 kg on the ligature blocks the trachea)

- Venous congestion : Jugular veins are blocked by the compression of the ligature which results in stoppage of the cerebral circulation. - Combined asphyxia and venous congestion : Most common cause - Cerebral anemia : Pressure on the large arteries of the neck produces cerebral anemia and instant coma - Reflex vagal inhibition leading to cardiac arrest - Fracture dislocation of cervical vertebrae.

Correct Answer. d

(129). The metabolic adaptation in alcoholic is all of the following except?

a. Lactic acidosis

b. High NAD+ level

c. Accumulation of fat in the liver

d. Decreased uric acid excretion

Solution. (b) High NAD+ level Ref.: Read the text below Sol :

- Oxidation of ethanol by alcohol dehydrogenase leads to excess production of NADH. - The increased (NADH/NAD+ ) ratio also causes increased (lactate/pyruvate) , resulting in hyperlacticacidemia, which decreases excretion of uric acid, aggravating gout.

Correct Answer. b

a. Glycolysis

b. TCA cycle

c. Uronic acid pathway

d. HMP pathway

Solution. (d) HMP pathway Ref.: Read the text below Sol : - HMP shunt path is most important path for production of NADPH. Fatty acid synthesis requires NADPH which is produced mainly by HMP shunt pathway. - HMP shunt pathway occurs in cytosol of the cell and it produces ribose and NADPH. - Ribose is utilized for nucleotide production and NADPH is utilized for reduction biosynthesis.

Correct Answer. d

(131). Which of the following enzymes is inhibited irreversibly by aspirin ?

a. Endoperoxidase

b. Cyclooxygenase

c. Lipoxygenase

d. Phospholipase C

Solution. (b) Cyclooxygenase Ref:Read the text below Sol:

- Cyclooxygenase is acetylated and inactivated by aspirin. - However, the hydrolytic removal of arachidonic acid from phospholipids, which is catalyzed by phospholipase A2, is the rate-limiting reaction for eicosanoid biosynthesis.

Correct Answer. b

(132). Xeroderma pigmentosa, is produced as a result of

a. DNA polymerase III

b. DNA polymerase I

c. DNA exonuclease

d. DNA ligase

Solution. (d) DNA ligase Ref:Read the text below Sol: Xeroderma pigmentosa is a defect of nucleotide excision repair. On exposure to UV light, there is pyrimidine- pyrimidine dimerisation. This has to be removed. The sequence of such a repair is as follows a. UV specific endonuclease/excinuclease: Cleaves approximately 25 to 28 nucleotide

long segment which includes the defective segment also. b. DNA polymerase:- i. It has two activities: It fragments the removed nucleotide fragment – exonuclease activity ii. DNA polymerase activity : For filling up the gap. c. DNA ligase : This helps in joining the newly synthesized strand with the parent strand.

Correct Answer. d

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 53/86 (133). Identify the correct sequence of actions in hybridisation technique/ southern blotting:- 1 PAGE 2. Blotting on nitrocellulose paper 3. Exposure to X-ray sheet 4. Addition of probe 5. Denaturation of ds DNA

a. 1 , 4 , 5 , 2 , 3

b. 1 , 2 , 3 , 5 , 4

c. 1 , 5 , 2 , 4 , 3

d. 1 , 2 , 5 , 4 , 3

Solution. (c) 1 , 5 , 2 , 4 , 3 Ref:Read the text below Sol: - Southern blotting is done for identifying the CDNA alone of our own interest,. First the various fragments of cDNA are electrophoresed. Then they are denatured with alkali. - Transferred to a nitro cellulose paper. - Probe annealed to the ss DNA fragments After through washing, paper is exposed to X-ray film - autoradiogram.

Correct Answer. c

(134). N-glycosylation of proteins occurs on which of the following amino acids ?

a. Asparagine

b. Aspartate

c. Lysine

d. Serine

Solution. (a) Asparagine Ref:Read the text below Sol: - After proteins are synthesized (translated), many are glycosylated in the lumen of the endoplasmic reticulum

and the Golgi complex. - Oligosaccharides are attached to proteins by N-glycosylation of asparagine side chains or by O-glycosylation of serine and threonine side chains. - The transfer of oligosaccharides is mediated by an activated lipid carrier, dolichol phosphate.

Correct Answer. a

(135). Which of the following enzymes may be targets of a new drug that specifically inhibits retroviral replication?

a. DNA-dependent DNA polymerase

b. Topoisomerase II

c. RNA-dependent DNA polymerase

d. RNA polymerase

Solution. (c) RNA-dependent DNA polymerase Ref:Read the text below Sol: - RNA-dependent DNA polymerase synthesizes DNA from an RNA template and is essential for the replication of

retroviruses (but not for cells) and therefore is a potential target of the new drug. - Topoisomerase II relaxes supercoiled DNA. - RNA polymerase synthesizes a primer fragment for DNA-dependent DNA polymerase. - DNA ligase anneals the Okazaki fragments on the lagging strand of DNA synthesis.

Correct Answer. c

a. Anneal Okazaki fragments

b. Relax supercoiled DNA

c. Degrade histone proteins

d. "Proofread" newly synthesized DNA

Solution. (c) RNA-dependent DNA polymerase Ref:Read the text below Sol: - RNA-dependent DNA polymerase synthesizes DNA from an RNA template and is essential for the replication of

Correct Answer. c

(137). Which one of the following activities is simultaneously stimulated by epinephrine in muscle and inhibited by epinephrine in the liver?

a. Fatty acid oxidation

b. Glycogenolysis

c. Cyclic AMP synthesis

d. Glycolysis

Solution. (d) Glycolysis Ref– Read the text below Sol: - Epinephrine stimulates both muscle and liver adenylate cyclase to produce cyclic AMP. - In the liver, the increased cyclic AMP levels activate a phosphatase that dephosphorylates fructose- 2,6- bisphosphate (F-2,6-BP) while deactivating a kinase that produces F-2,6-BP. - Thus, F-2,6-BP levels are decreased and phosphofructokinase activity is decreased. In liver and muscle, F-2,6-BP

is the major allosteric activator of phosphofructokinase. In skeletal muscle, however, the kinase responsible for the synthesis of F-2,6-BP is activated, not inhibited, by cyclic AMP. - Thus, muscle sees an increase in glycolysis following epinephrine stimulation, while the liver experiences a decrease in glycolytic activity. - In both tissues, glycogen phosphorylase is activated and glycogenolysis occurs. - Under these conditions, glucose is utilized in muscle for ATP production relative to contractile activity, while the liver produces glucose for export to the blood.

Correct Answer. d

a. Hanging

b. Infanticide

c. Drowning

d. OP compound poisoning

Solution. (b) Infanticide Ref.: Narayan Redy’s - 208. Sol : - Hydrostatic test (floatation test; Raygat’s test) is a test done to confirm whether the lungs tested are from a respired newborn or not, i.e. this test is used to demonstrate the signs of livebirth.

Correct Answer. b

(139). Greenish blue urine occurs in the following condition

a. Rifampicin therapy

b. Naphthalene poisoning

c. Phenol poisoning

d. Methemoglobinemia.

Solution. (c) Phenol poisoning Ref.: Narayan Reddy’s - 262. Sol :

- Phenol is metabolized by the kidneys causing renal damage with excretion of scanty, dark colored urine, which on standing turns smoky green – Carboluria. - It contains phenolic metabolites like hydroquinone and pyrocatechol.

Correct Answer. c

(140). Hesitation marks are characteristic of

a. Accident injury in young child

b. Suicidal wounds

c. Homicidal wounds inflicted by physically weak man

d. Accidental injury

Solution. (b) Suicidal wounds Ref.: Narayan Reddy’s - 101. Sol : - Hesitation marks or tentative cuts are cuts which are multiple, small and superficial often involving only the skin and are seen at the beginning of the incised wound. It is characteristic of suicidal wound.

Correct Answer. b

a. 6 hours

b. 12 hours

c. 24 hours

d. 48 hours

Solution. (a) 6 hours Ref.: Narayan Reddy’s - 191. Sol : - The spermatozoa deposited in the vagina lose motility within 1 hour and at the end of 6 hours, no motile sperms are found. - Sperms may be recovered up to 24 hours from the vaginal.

Correct Answer. a

(142). In a charred body which of the following is a useful means of establishing identity ?

a. Stature

b. Comparison of postmortem X-rays with dental records

c. Scar marks

d. Skeletal fractures

Solution. (b) Comparison of postmortem X-rays with dental records Ref.: Narayan Reddy’s FMT- 56. Sol : - In a charred body, stature may be less by several centimeters and weight loss may be up to 60% - Moles, scars and tattoo marks are usually destroyed.

- Dental identification (forensic odontology) depends mainly upon comparison between records of the missing persons and the findings in the charred bodies in relation to - Restorative work - Unusual features - Comparison of ante mortem with postmortem X-rays.

Correct Answer. b

(143). Which one of the following features listed below is most suggestive of antemortem hanging ?

a. Salivary dribbling

b. Congestion of lungs

c. Ligature mark

d. Petechial hemorrhages

Solution. (a) Salivary dribbling Ref.: Narayan Reddy’s - 157.

Sol : - Salivary dribbling is most suggestive of antemortem hanging

Correct Answer. a

a. Potassium nitrate

b. Sulphur

c. Magnesium nitrate

d. Charcoal

Solution. (c) Magnesium nitrate Ref:Read the text below Sol: Gun powder are of two type : (A) Black powder - It consist of potassium nitrate 75%, sulphur 10% & Charcoal 15%. (B) Smoke less powder - It consists of nitro cellulose (Gun cotton), or nitroglycerine & nitro cellulose (Double base) or nitroglycerine, nitro cellulose and nitroguanadine (Triple base)

Correct Answer. c

(145). Which of the following statements regarding electricity is correct?

a. A.C. is more dangerous than D.C.

b. D.C. is more dangerous than A.C.

c. Both are equally dangerous

d. Which is more dangerous depends on the fact if one is standing on a wet or a dry surface

Solution. (a) A.C. is more dangerous than D.C. Ref: Read the text below Sol:

- A.C. is about 4-5 times as dangerous as D.C. - It is because of the tetanoid contractions effected by the A.C. - An A.C. of 70-80 mA may be lethal, whereas a D.C. with an intensity of 250 mA may be tolerated without damage.

Correct Answer. a

(146). Which of the following events occurs during formation of phosphoenolpyruvate from pyruvate during gluconeogenesis?

a. CO2 is consumed

b. Inorganic phosphate is consumed

c. Acetyl CoA is utilized

d. ATP is generated

Solution. (a) CO2 is consumed Ref– Read the text below Sol: In the formation of phosphoenolpyruvate during gluconeogenesis, oxaloacetate is an intermediate. In the first step, catalyzed by pyruvate carboxylase, pyruvate is carboxylated with the utilization of one high-energy ATP phosphate bond: Pyruvate + ATP + CO2

-oxaloacetate + ADP + Pi - In the second step, catalyzed by phosphoenolpyruvate carboxykinase, a high-energy phosphate bond of GTP drives the decarboxylation of oxaloacetate: Oxaloacetate + GTP -phosphoenolpyruvate +GDP+CO2 - In contrast to gluconeogenesis, the formation of pyruvate from phosphoenolpyruvate during glycolysis requires only pyruvate kinase, and ATP is made.

Correct Answer. a

a. Fumarate

b. Pyruvate

c. Oxaloacetate

d. Succinyl CoA

Solution. (d) Succinyl CoA Ref– Read the text below Sol: - The final thiolytic cleavage in - --oxidation of odd-chain fatty acids yields propionyl CoA. - Propionyl CoA is also formed during the breakdown of methionine and isoleucine.It is carboxylated to form D- methylmalonyl CoA, which is in equilibrium with 1-methylmalonyl CoA.

- Valine forms methylmalonyl CoA during its degradation. The 1-isomer of methylmalonyl CoA is converted to succinyl CoA through the action of the B12 coenzyme–containing methylmalonyl CoA mutase. - Thus, succinyl CoA serves as the entry point into the citric acid cycle for three amino acids and the last three carbons of odd-chain fatty acids. - The amino acids and fatty acid carbons introduced in this manner may either be catabolized in the cycle for energy production or utilized for gluconeogenesis.

Correct Answer. d

(148). Cholera toxin causes massive and often fatal diarrhea by

a. Inactivating Gi protein

b. Irreversibly activating adenylate cyclase

c. Locking Gs protein into an inactive form

d. Rapidly hydrolyzing G protein GTP to GDP

Solution. (b) Irreversibly activating adenylate cyclase Ref– Read the text below Sol: - Cholera toxin is an 87-kD protein produced by Vibrio cholerae, a gram-negative bacterium. - The toxin enters intestinal mucosal cells by binding to GM1 ganglioside. - It interacts with Gs protein, which stimulates adenylate cyclase. By ADP-ribosylation of Gs, the toxin blocks its capacity to hydrolyze bound GTP to GDP. - Thus, the G protein is locked in an active form and adenylate cyclase stays irreversibly activated. Under normal

conditions, inactivated G protein contains GDP, which is produced by a phosphatase catalyzing the hydrolysis of GTP to GDP. - When GDP is so bound to the G protein, the adenylate cyclase is inactive. Upon hormone binding to the receptor, GTP is exchanged for GDP and the G protein is in an active state, allowing adenylate cyclase to produce cyclic AMP. - Because cholera toxin prevents the hydrolysis of GTP to GDP, the adenylate cyclase remains in an irreversibly active state, continuously producing cyclic AMP in the intestinal mucosal cells. This leads to a massive loss of body fluid into the intestine within a few hours.

Correct Answer. b

a. Liver

b. Muscle

c. Hepatocytes

d. Brain

Solution. (b) Muscle Ref– Read the text below Sol: - Muscle cells are the only cells listed that are capable of utilizing all the energy sources available—glucose, fatty acids, and, during fasting, ketone bodies.

- Mitochondria are required for metabolism of fatty acids and ketone bodies. Since red blood cells (erythrocytes) do not contain mitochondria, no utilization of these energy sources is possible. - Although the brain may utilize glucose and ketone bodies,fatty acids cannot cross the blood-brain barrier. - Hepatocytes (liver cells) are the sites of ketone body production, but the mitochondrial enzyme necessary for utilization of ketone bodies is not present in hepatocytes.

Correct Answer. b

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 60/86 (150). There is but a single enzyme-catalyzed reaction in the human body known to generate carbon monoxide (CO) as one of its products.Which of the following enzymes represents the one that catalyzes this CO-producing reaction?

a. Biliverdin reductase

b. Coproporphyrinogen oxidase

c. Heme oxygenase

d. Protoporphyrinogen oxidase

Solution. (c) Heme oxygenase Ref– Read the text below Sol: Heme is oxidized, with the heme ring being opened by the endoplasmic reticulum enzyme, heme oxygenase (see Fig). - The oxidation step requires heme as a substrate, and any hemin (Fe3+) is reduced to heme (Fe2+) prior to oxidation by heme oxygenase. - The oxidation occurs on a specific carbon producing the linear tetrapyrrole biliverdin, ferric iron (Fe3+), and CO. This is the only reaction in the body that is known to produce CO. - Most of the CO is excreted through the lungs, with the result that the CO content of expired air is a direct measure of the activity of heme oxidase in an individual.

Correct Answer. c

a. ACTH

b. Follicle-stimulating hormone

c. Growth hormone

d. Prolactin

Solution. (b) Follicle-stimulating hormone Ref– Read the text below Sol: - The pituitary gonadotrophs are cells that secrete the gonadotrophic peptide hormones,follicle-stimulating hormone (FSH), and leutinizing hormone (LH). FSH acts on Sertoli cells of the testis inducing androgen-binding

protein synthesis, which maintains a high concentration of testosterone in tubules increasing spermatogenesis. - In the ovary, FSH stimulates aromatase activity in the granulose cells stimulating ovum maturation and estradiol production. - Pituitary orticotrophs are the ACTH-secreting cells, lactotrophs secrete prolactin and thyrotrophs secrete thyroid-stimulating hormone.

Correct Answer. b

(152). The rate-limiting step in glycolysis occurs at the step catalyzed by which of the following enzymes?

a. Glyceraldehyde-3-phosphate dehydrogenase

b. 6-phosphofructo-1 kinase, PFK-1

c. 6-PFK-2

d. Phosphoglycerate kinase

Solution. (b) 6-phosphofructo-1 kinase, PFK-1 Ref– Read the text below Sol: - There are three reactions of glycolysis that are thermodynamically irreversible. These are the hexokinase (glucokinase), PFK-1, and PKcatalyzed reactions. Reactions that are essentially irreversible in most metabolic pathways are subject to complex regulatory controls and represent rate-limiting steps in the pathway. - The primary site of regulation of glycolysis occurs at the level of the PFK-1-catalyzed step.Hence, this reaction is the rate-limiting step in glycolysis. PFK-1 is subject to allosteric control by numerous compounds. - Citrate and ATP inhibit the activity of PFK-1, while AMP and fructose 2,6-bisphosphate (F2,6-BP) activate the enzyme. The principal control of PFK-1 activity is exerted by alterations in the level of F2,6-BP. This compound is synthesized from fructose-6-phosphate by the bifunctional enzyme, PFK-2/fructose-2,6-bisphosphatase (PFK- 2/F2,6-BPase). - PFK-2 contains two catalytic domains, one a kinase and the other a phosphatase, the activities of which are affected by the state of phosphorylation. The phosphatase domain is active when the enzyme is phosphorylated and converts F2,6-BP back to F6-P, thereby reducing the levels of this powerful activator of PFK-1. - Thus, although the activity f PFK-2 will determine the rate of activity of PFK-1, it is itself not the rate-limiting enzyme in glycolysis.

Correct Answer. b

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 62/86 (153). The statin class of drugs that are currently used to control hypercholesterolemia function to lower circulating levels of cholesterol by which of the following mechanisms?

a. Increasing the elimination of bile acids, leading to increased diversion of cholesterol into bile acid production

b. Increasing the synthesis of apolipoprotein B-100 (apo B100), resulting in increased elimination of cholesterol through the action of

low-density lipoprotein (LDL) uptake by the liver

c. Inhibiting the interaction of LDLs with the hepatic LDL receptor

d. Inhibiting the rate-limiting step in cholesterol biosynthesis

Solution. (d) Inhibiting the rate-limiting step in cholesterol biosynthesis Ref– Read the text below Sol: - The statin class of cholesterol-lowering drugs all exert their effects on the activity of HMG-CoA reductase.

- This enzyme carries out the rate-limiting step in cholesterol biosynthesis.The cholestyramine-based resins are used therapeutically to bind up intestinal bile salts,which increases the excretion of bile. - The net effect of increased bile excretion is increased diversion of cholesterol into bile acid synthesis, thereby lowering circulating cholesterol levels.

Correct Answer. d

(154). Which of the following represents the enzyme deficiency that leads to “essential fructosuria”?

a. Fructose-1-phosphate aldolase (aldolase B)

b. Fructose-1,6-bisphosphate aldolase (aldolase A)

c. Fructokinase

d. Hexokinase

Solution. (c) Fructokinase Ref– Read the text below Sol: - Essential fructosuria is an autosomal recessive disorder manifesting benign asymptomology due to a lack of fructokinase. - The principal clinical signs of essential fructosuria are hyper-fructosemia and fructosuria.Deficiency in aldolase B results in the clinically severe disorder, hereditary fructose intolerance. - Symptoms include severe hypoglycemia and vomiting after ingestion of fructose. Prolonged fructose ingestion by infants with this disorder will lead to poor feeding, hepatomegaly, vomiting, jaundice, and eventually hepatic failure and death. - Aldolase A deficiency is very rare and has only been observed associated with erythrocytes and fibroblasts and leads to hemolytic anemia.Hexokinase deficiency is also rare and associated with erythrocyte disfuction leading to chronic hemolytic anemia.

Correct Answer. c

a. Beta-hydroxybutyrate dehydrogenase

b. Hydroxymethylglutaryl-CoA-lyase

c. Hydroxymethylglutaryl-CoA-synthetase

d. Succinyl-CoA-acetoacetate-CoAtransferase

Solution. (d) Succinyl-CoA-acetoacetate-CoAtransferase Ref– Read the text below Sol: - Ketogenesis occurs in the liver from acetyl- CoA during high rates of fatty acid oxidation and during early starvation. - The principal ketone bodies are acetoacetate and betahydroxybutyrate, which are reversibly synthesized in a reaction catalyzed by betahydroxybutyrate dehydrogenase. - The liver delivers beta-hydroxybutyrate to the circulation where it is taken up by non-heptatic tissue for use

as an oxidizable fuel. - The brain will derive much of its energy from ketone body oxidation during fasting and starvation. Within extrahepatic tissues, beta-hydroxybutyrate is converted to acetoacetate by beta-hydroxybutyrate dehydrogenase. - Acetoacetate is reactivated to acetoacetyl-CoA in a reaction catalyzed by succinyl-CoA-acetoacetate-CoA- transferase (also called acetoacetate: succinyl-CoA-CoA transferase), which uses succinyl-CoA as the source of CoA. - This enzyme is not present in hepatocytes. The acetoacetyl-CoA is then converted to 2 moles of acetyl-CoA by the thiolase reaction of fatty acid oxidation.

Correct Answer. d

(156). Contre-coup injuries are seen in?

a. Brain

b. Heart

c. Liver

d. Pancreas

Solution. (a) Brain. Ref– Read the text below Sol: - In head injury, a coup injury occurs under the site of impact with an object, and a contrecoup injury occurs on the side opposite the area that was impacted. - Coup and contrecoup injury is associated with cerebral contusion, a type of traumatic brain injury in which the brain is bruised.

Correct Answer. a

(157). Study of death is known as:

a. Thanatology

b. Triochology

c. Teratology

d. Chieloscopy

Solution. (a) Thanatology. References: Parikh’s -3.1 Sol: Thanatology. The word comes from Thanatos, who in Greek mythology was the personification of Death. Other studies:- - Entomology of cadaver-for time determination since death - Dactylography-study of fingerprint for identification - Chieloscopy – study of lip prints for identification - Bertillon system/anthropometry identification of adult > 21 years of age

Correct Answer. a

(158). Not a part of informed consent is:

a. Discussion of pertinent information

b. Patient’s agreement to the plan of care

c. Freedom from rejection.

d. Concealed information.

Solution. (d) Concealed information. References: Read the text below Sol: Informed consent: Informed consent implies an understanding by the patient of: - The nature of the condition. - The nature of the proposed treatment or procedure. - The risks and benefits involved in both the proposed and alternative procedures. - The potential risks of not receiving treatment. - The relative chances of success or failure of both procedures, so he may accept or reject the procedure.

Correct Answer. d

(159). Hydrocution is ?

a. Electric shock in water

b. Cold water submersion

c. Submersion in boiling water

d. None

Solution. (b) Cold water submersion References: Read the text below Sol: Hydrocution

- Also called- immersion syndrome or vagal inhibition - It is a type of atypical drowning - Mechanism –sudden dyspnoea and vagal inhibition occur as a result – heart stops immediately. In some cases sudden ventricular fibrillation has also been reported.

Correct Answer. b

a. Lead

b. Hg

c. As

d. Sulphur

Solution. (c) As References: Sharma 230 Sol: This is a classical case presentation of Arsenic poisoning Metallic form – Non poisonous, insoluble-not digested

- Salts with O2 , Cl-toxic [oxidation occur in air] - Salts interfere with cell metabolism by combining with sulphydryl enzyme Acute poisoning – occur within 15-30 min - Metalic taste,slight garlicky smell,dry mouth – N/V,colicky abdominal pain – diarrhea,rice water stool (may be bloody),dilatation of capillary and damage,intense thirst, suppressed urine - Death d/t hypovolumic stock - Sensory motor (mixed) peripheral neuropathy within few hours of severe cases-respiratory failure, and quadriplegia Chronic Small quantity slowly and repeated Dyspepsia vomiting, purging, bloody motions, cutaneous blister More marked symp of neuritis, severe muscular pain - 1st stage predominant git symptoms - 2-4 weeks-Mees line – white transverse band in nail plate [Pterygium-longitudinal/vertical band] which may remain upto 1 year *mee’s line seen in 2nd stage of As poisoning] - 3rd stage absent knee jerk and bone marrow depression - 4th stage peripheral neuropathy megaloblastic anaemia, foot drop. Fatal dose-180 mg of As trioxide Treatment - Lavage, should be with pain water first then fresh hydrated ferric oxide - Activated charcoal if ferric oxide not avail - Exchange transfusion, hemodilysis - Chelation-BAL Tests = Reinsch’s test, Marsh’s test, Gutzeit test

Correct Answer. c

(161). Falanga is:

a. Beating the soles and stick with blunt object

b. Immersing head in water

c. Twisting of ears.

d. Trauma to eyes.

Solution. (a) Beating the soles and stick with blunt object References: Read the text below Sol:

FALANGNA a) Falanga is beating the soles and stick with blunt object b) Common in police beating

Correct Answer. a

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 66/86 (162). A 42-year-old man sustains severe injuries in an automobile accident and is admitted to the intensive care unit.Examination of a peripheral blood smear on the 3rd day of admission reveals helmet cells, schistocytes, and decreased platelets. Which of the following is most strongly suggested by these findings?

a. Autoimmune hemolysis

b. Disseminated intravascular coagulation (DIC)

c. Hereditary spherocytosis

d. Megaloblastic anemia

Solution. (b) Disseminated intravascular coagulation (DIC) References: Read the text below Sol: - The findings suggest disseminated intravascular coagulation (DIC), which is a feared complication of many

other disorders, such as obstetrical catastrophes, metastatic cancer, massive trauma, and bacterial sepsis - The basic defect in DIC is a coagulopathy characterized by bleeding from mucosal surfaces, thrombocytopenia, prolonged PT and PTT, decreased fibrinogen level, and elevated fibrin split products. - Helmet cells and schistocytes (fragmented red blood cells) are seen on peripheral blood smear.

Correct Answer. b

(163). A 57-year-old fisherman with a history of alcoholism is hospitalized in with a 1-day history of severe, watery diarrhea after eating several raw oysters. He is badly dehydrated on admission, and within 12 hours, he becomes severely hypotensive and dies. Which of the following pathogens is the most likely cause of this man's death?

a. Enterotoxigenic E. coli

b. Providencia stuartii

c. Vibrio cholerae

d. Vibrio vulnificus

Solution. (d) Vibrio vulnificus References: Read the text below Sol: - Vibrio vulnificus is an extremely invasive organism, producing a septicemia in patients after eating raw

shellfish, or causing wound infections, cellulitis, fasciitis, and myositis after exposure to seawater or after cleaning shellfish. - Patients at high risk for septicemia include those with liver disease, congestive heart failure, diabetes mellitus, renal failure, hemochromatosis, and immunosuppression

Correct Answer. d

(164). A baby is born with a flat facial profile, prominent epicanthal folds,and simian crease. She vomits when fed, and upper GI studies demonstrate a "double bubble" in the upper abdomen. Which of the following cardiovascular abnormalities might this child also have?

a. Atrial septal defect

b. Berry aneurysm

c. Coarctation of the aorta

d. Endocardial cushion defect

Solution. (d) Endocardial cushion defect References: Read the text below Sol: - The disease is Down syndrome (trisomy 21).

- In addition to mental retardation and the characteristic physical findings described in the question stem, duodenal atresia is fairly common, as evidenced by the "double bubble" sign on x-ray. - These children are also likely to have various cardiac anomalies; endocardial cushion defect is the most common.

Correct Answer. d

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 67/86 (165). Imprisonment for a term extending to seven years and also fine as punishment for voluntarily causing grievous hurt : -

a. S. 321, I.P.C

b. S. 322, I.P.C

c. S. 325, I.P.C

d. S. 324, I.P.C

Solution. (c) S. 325, I.P.C Ref: Read the text below Sol: S.325, I.P.C - Punishment for voluntarily causing grievous hurt: Imprisonment for a term extending to seven years and also fine

Correct Answer. c

(166). Close shot is applied when the victim is within

a. 2 to 3 cm.

b. 3 to 5 cm.

c. 5 to 8 cm.

d. 8 to 9 cm.

Solution. (c) 5 to 8 cm. Ref: Read the text below Sol:

CLOSE SHOT - This term is applied when the victim is within the range of the flame, i.e. 5 to 8 cm. The term ‘point blank’ is used when the range is very close to or in contact with the surface of the skin.

Correct Answer. c

(167). Which of the following statements reflects the process by which the telomeric ends of chromosomes are replicated?

a. A unique RNA molecule serves as the primer for synthesis.

b. A unique RNA molecule serves as the template for synthesis.

c. Short template-independent block of DNA are ligated to the ends using a 5’-5’ bond.

d. Telomeres are replicated in a templeindependent process.

Solution. (b) A unique RNA molecule serves as the template for synthesis. Ref– Read the text below Sol:

- The telomeric end on the lagging strand of a replicating eukaryotic chromosomes are synthesized by an enzymatic activity termed telomerase. - None of the other mechanisms represent the accurate replication of telomeres.

Correct Answer. b

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 68/86 (168). In carrying out an assay using cultured hepatocytes, you find that addition of hemin (Fe3+heme) does not have the expected consequence of reduced protoporphyrin IX synthesis. This result suggests that your hepatocytes harbor a mutant form of one of the heme-regulated enzymes of porphyrin biosynthesis. Which of the following represents the likely enzyme?

a. ALA dehydratase

b. ALA synthase

c. Ferrochelatase

d. Heme oxygenase

Solution. (b) ALA synthase Ref– Read the text below Sol: - In the liver, hemin acts as a feedback inhibitor on ALAsynthase reducing its activity.

- In addition, hemin acts to repress transport of ALAsynthase into the mitochondria as well as repressing synthesis of the enzyme. Therefore,a continued synthesis of protoporphyrin IX in the presence of hemin indicates that ALA synthase is active but nonresponsive to the inhibitory action of hemin. - None of the other enzymes of heme biosynthesis

Correct Answer. b

(169). Oxidative degradation of acetyl coenzyme A (CoA) in the citric acid cycle gives a net yield of which of the following chemicals?

a. Flavin adenine dinucleotide (FAD+)

b. Nicotinamide adenine dinucleotide (NAD+)

c. Adenosine triphosphate (ATP)

d. Carbon dioxide (CO2)

Solution. (d) Carbon dioxide (CO2) Ref– Read the text below Sol: The net result of the citric acid cycle’s oxidation of acetyl CoA is shown below: acetyl CoA + FAD + 3 NAD + + GDP + 2 H2O +Pi 2 CO2 +CoA +FADH2 +3 NADH +GTP +2 H + - The cycle produces reducing equivalents (NADH, FADH2) and carbon dioxide directly, but not ATP. The reducing equivalents are used to produce ATP by mitochondrial oxidative phosphorylation.

- Two carbon atoms enter the cycle as acetyl CoA, with an immediate loss of CoA as citrate is formed from oxaloacetate. Two carbon atoms leave the cycle at the level of isocitrate dehydrogenase and α-ketoglutarate dehydrogenase. - The two carbons that leave as CO2 are actually not the original acetyl CoA carbons. Two NAD+ molecules are reduced by isocitrate dehydrogenase and then α- ketoglutarate dehydrogenase. - One FAD is reduced during the oxidation of succinate, and one NAD is reduced when malate is oxidized. GTP is formed from GDP by utilization of the high-energy thioester linkage of succinyl CoA.

Correct Answer. d

(170). Guanosine triphosphate (GTP) is required by which of the following steps in protein synthesis?

a. Aminoacyl-tRNA synthetase activation of amino acids

b. Attachment of ribosomes to endoplasmic reticulum

c. Translocation of tRNA–nascent protein complex from A to P sites

d. Attachment of mRNA to ribosomes

Solution. (c) Translocation of tRNA–nascent protein complex from A to P sites Ref– Read the text below Sol: - Two molecules of GTP are used in the formation of each peptide bond on the ribosome.

- In the elongation cycle, binding of aminoacyl-tRNA delivered by EF-Tu to the A site requires hydrolysis of one GTP. Peptide bond formation then occurs. - Translocation of the nascent peptide chain on tRNA to the P site requires hydrolysis of a second GTP. - The activation of amino acids with aminoacyl-tRNA synthetase requires hydrolysis of ATP to AMP plus PPi.

Correct Answer. c

(171). A potent inhibitor of protein synthesis that acts as an analogue of aminoacyl-tRNA is

a. Streptomycin

b. Nalidixic acid

c. Rifampicin

d. Puromycin

Solution. (d) Puromycin Ref– Read the text below Sol: - Puromycin is virtually identical in structure to the 3′-terminal end of tyrosinyl-tRNA. In both eukaryotic and prokaryotic cells, it is accepted as a tyrosinyl-tRNA analogue.

- As such, it is incorporated into the carboxy-terminal position of a peptide at the aminoacyl (A) site on ribosomes, causing premature release of the nascent polypeptide. - Thus,puromycin inhibits protein synthesis in both human and bacterial cells.Streptomycin, like tetracycline and chloramphenicol, inhibits ribosomal activity. - Rifampicin is an inhibitor of bacterial DNA-dependent RNA polymerase.

Correct Answer. d

(172). Which one of the following hormones is derived most completely from tyrosine?

a. Glucagon

b. Thyroxine

c. Insulin

d. Prostaglandins

Solution. (b) Thyroxine Ref– Read the text below Sol: - Two of the major hormones are derived from the amino acid tyrosine: the adrenal hormone epinephrine and

the thyroid hormone thyroxine (tetraiodothyronine). - Epinephrine is the catabolic antagonist of insulin, a polypeptide hormone, and is similar in action to glucagon, a liver-specific polypeptide hormone. - Thyroxine is important in governing the basal metabolic rate.

Correct Answer. b

b. Sphingomyelin

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 74/86 Solution. (b) Sphingomyelin Ref– Read the text below Sol: - The most common sphingolipid in mammals is sphingomyelin.Ceramide, the basic structure from which all sphingolipids are derived, is composed of the 18-carbon sphingosine connected via its amino group to a fatty acid by an amide linkage. - The fatty acid is usually long-chain (18 to 26 carbons) and is saturated or monosaturated. Except for the lack of the glycerol backbone, sphingolipids are quite similar in structure and physical properties to the phospholipids phosphatidyl choline and phosphatidyl ethanolamine. - Either phosphoryl choline or phosphoryl ethanolamine is the head group attached to ceramide. - If a neutral sugar residue is the polar head group attached to ceramide, a cerebroside is formed. If oligosaccharide head groups containing sialic acid are used, gangliosides are formed. - All sphingolipids are important membrane constituents.

(174). Which of the following steps in the biosynthesis of cholesterol is thought to be rate- controlling and the locus of metabolic regulation?

a. Geranyl pyrophosphate →farnesyl pyrophosphate

b. Squalene →lanosterol

c. Lanosterol →cholesterol

d. 3-hydroxy-3-methylglutaryl CoA →mevalonic acid

Solution. (d) 3-hydroxy-3-methylglutaryl CoA →mevalonic acid Ref– Read the text below Sol: - Regulation of cholesterol metabolism is by definition exerted at the “committed” and rate-controlling step. This is the reaction catalyzed by 3-hydroxy-3-methylglutaryl CoA reductase. - Reductase activity is reduced by fasting and by cholesterol feeding and thus provides effective feedback control of cholesterol metabolism. The statin class of drugs act at this site.

Correct Answer. d

(175). Which of the following statements correctly describes the enzyme thiokinase?

a. It yields acetyl CoA as a product

b. It yields ADP as a product

c. It yields CoA as a product

d. It forms CoA thioesters as a product

Solution. (d) It forms CoA thioesters as a product Ref– Read the text below Sol: - Fatty acids must be activated before being oxidized. In this process, they are linked to CoA in a reaction

catalyzed by thiokinase (also known as acyl CoA synthetase). = - ATP is hydrolyzed to AMP plus pyrophosphate in this reaction. - In contrast, the enzyme thiolase cleaves off acetyl CoA units from β-ketoacyl CoA, while it forms thioesters during βoxidation.

Correct Answer. d

(176). For every 2 mol of free glycerol released by lipolysis of triacylglycerides in adipose tissue

a. 2 mol of triacylglycerides is released

b. 2 mol of free fatty acids is released

c. 1 mol of glucose can be synthesized in gluconeogenesis

d. 1 mol of triacylglyceride is released

Solution. (c} 1 mol of glucose can be synthesized in gluconeogenesis Ref– Read the text below Sol: - During lipolysis, triglycerides are split into three free fatty acids and glycerol.The free fatty acids, as well as the free glycerol, diffuse into the bloodstream, where they are circulated throughout the body. - The free fatty acids are used as an energy source for many tissues, primarily muscle. The free glycerol that is

released cannot be phosphorylated back to glycerol-3-phosphate in the adipose tissue because it lacks glycerol kinase. - However, the free glycerol released in lipolysis is taken up by the liver, where it can be phosphorylated to glycerol-3-phosphate. The phosphorylated glycerol can enter glycolysis or gluconeogensesis at the level of triose phosphates. - If gluconeogenesis occurs, for every 2 mol of glycerol-3-phosphate, 1 mol of glucose can be synthesized.

Correct Answer. c

a. Connects Okazaki fragments

b. Requires ATP for activity

c. Acts on DNA-RNA hybrids as a substrate

d. Is required to form a DNA double helix

Solution. (b) Requires ATP for activity Ref:Read the text below Sol: - DNA topoisomerase II is an ATP-dependent enzyme that introduces negative supercoils into the helix preceding

replication forks. - This relieves positive supercoils at the replication fork. - Etoposide and teniposide are topoisomerase II inhibitors that are used in the treatment of lymphomas, Kaposi sarcoma, and cancers of the lung, testis, and breast.

Correct Answer. b

(178). Discoloration or staining of the skin and organs, known as post-mortem lividity occurs in the dead body due to accumulation of fluid blood in the dependent parts of the body. The process fully developed and fixed in about

a. 1-3 hours.

b. 3-6hours.

c. 6-12 hours.

d. 12-24 hours.

Solution. (c) 6-12 hours Ref– Read the text below Sol: EARLY SIGN OF DEATH - Discoloration or staining of the skin and organs, known as post-mortem lividity occurs in the dead body due to accumulation of fluid blood in the dependent parts of the body. - The process commences within an hour after death and fully developed and fixed in about 6-12 hours. - Post-mortem lividity is very important in medico-legal cases. It is a reliable sign of death, it gives information about the position of the body at the time of death and any disturbance caused in the dead body as usually happens in homicidal cases. - It also helps to estimate the time since death.

Correct Answer. c

(179). Rule of Hasse is used to determine?

a. The age of fetus

b. Height of an adult

c. Race of a person

d. Identification

Solution. (a) The age of fetus Ref– Rea d the text below Sol: HASSE’S RULE –assessment of age of fetus Crown-Heel Length : CHL

- CHL in the first half of pregnancy is the number of lunar months x 4. The CHL of a 4 month fetus is 16cm : 4x4=16 cm - From the end of 20 weeks in the second half of pregnancy, CHL in cm is the result of multiplication of the number of lunar months at the time of the assessment by 5. The CHL of an 8-month fetus is 40 cm: 8x 5 =40 cm

Correct Answer. a

a. Shot gun

b. Revolver

c. Pistol

d. Rifle

Solution. (c) Pistol Ref– Read the text below Sol: PISTOL - A pistol automatically ejects the empty case and reloads a round from a magazine into the chamber after the preceding round has been

fired, feature differentiating it from a revolver. - The advantage of pistol is the use of recoil generated by the fired cartridge to eject the empty cartridge case, load the next cartridge, and cock the hammer. This is more conducive to firing multiple shots, so many are designed to carry 15 to 19 rounds. - Disadvantages include a more complicated mechanism, require more practice to use, and cartridge cases must be short to work well. Revolver cartridges are more powerful than semiautomatic cartridges for this last reason.

Correct Answer. c

(181). Basophilic stippling is seen in:

a. RBC

b. WBC

c. Neutrophils

d. Basophiles

Solution. (a) RBC Ref– Read the text below Sol: Basophilic stippling of erythrocytes (BSE) - Basophilic stippling of erythrocytes represents the spontaneous aggregation of ribosomal RNA in the cytoplasm of erythrocytes. - These aggregates stain, and hence are visible, with routine hematology stains. BSE may be seen as a feature of regenerative anemia,BSE can also occur in the absence of anemia with lead poisoning. - In this situation, stippling results from the poisoning of the 5' nucleotidase normally responsible for degrading RNA, and hence may be seen in older, smaller red cells as well as nucleated red blood cells.

Correct Answer. a

(182). Suspended animation occurs after

a. Lead poisoning.

b. Cyanide poisoning

c. Burns

d. Drowning

Solution. (d) Drowning Ref– Read the text below Sol: Suspended animation occurs in: - Electrocution Cholera

- Drowning Deep shock - New born Barbiturates / opioids overdose - After anaesthesia Insanity - Cerebral concussion - Heat stroke

Correct Answer. d

a. High court

b. Supreme court

c. Session court

d. Assistant session court

Solution. (a) High court Ref– Read the text below Sol - Sentence of death is finally passed high court - When the Court of Session passes a sentence of death, the proceedings shall be submitted to the High Court, and the sentence shall not be executed unless it is confirmed by the High Court. - The Court passing the sentence shall commit the convicted person to jail custody under a warrant.

Correct Answer. a

(184). The Court of a Chief Judicial Magistrate may pass any imprisonment for a term less than

a. 5 yrs.

b. 7 yrs.

c. 9 yrs.

d. 10 yrs.

Solution. (b) 7 yrs. Ref– Read the text below Sol: The Court of a Chief Judicial Magistrate may pass any sentence authorised by law except a sentence of death or of imprisonment for life or of imprisonment for a term exceeding seven years

Correct Answer. b

(185). A man is said to commit "rape" who has sexual intercourse with a woman under all of the following circumstances except;

a. With her consent, when the man knows that he is not her husband.

b. By a man with his own wife, the wife not being under fifteen years of age

c. If consent is given because she believes that he is another man to whom she is or believes herself to be lawfully married.

d. With her consent, when she is under sixteen years of age.

Solution. (b) By a man with his own wife, the wife not being under fifteen years of age Ref– Read the text below Sol: Rape A man is said to commit "rape" who, except in the case hereinafter excepted, has sexual intercourse with a woman under circumstances falling under any of the six following descriptions – - Firstly – Against her will. - Secondly – Without her consent. - Thirdly – With her consent, when her consent has been obtained by putting her or any person in whom she

is interested in fear of death or of hurt. - Fourthly – With her consent, when the man knows that he is not her husband, and that her consent is given because she believes that he is another man to whom she is or believes herself to be lawfully married. - Fifthly – With her consent, when, at the time of giving such consent, by reason of unsoundness of mind or intoxication or the administration by him personally or through another of any stupefying or unwholesome substance, she is unable to understand the nature and consequences of that to which she gives consent. - Sixthly – With or without her consent, when she is under sixteen years of age. - Explanation.-Penetration is sufficient to constitute the sexual intercourse necessary to the offence of rape. - Exception.-Sexual intercourse by a man with his own wife, the wife not being under fifteen years of age, is not rape.

Correct Answer. b

a. Putrefaction

b. Adipocere

c. Mummification

d. Rigor mortis

Solution. (c) Mummification Ref– Read the text below Sol: - Putrefaction or decomposition is the last stage of the body to convert from organic to inorganic state and is an absolute sign of death. - The putrefaction is brought about by autolysis i.e. due to rise in die enzyme levels, the tissues get softened and

commences three to four hours after death. - Putrefaction also is brought about by the bacteria in such atmospheric conditions like warmth, humidity, air etc. - Adipocere results form hydrogenation of unsaturated body fats into firmer fats and their hydrolysis into fatty acids. It is a yellowish white, greasy, wax like substance with a rancid smell. - Mummification is characterised by dehydration of body tissues and viscera after death.

Correct Answer. c

(187). Pearson's formulae is to measure the length of long bone and multiply it with a given factor, and then adding a fixed factor. Multiplying factor for fibula is : -

a. 6 to 6.3

b. 4.2 to 4.5

c. 4.4.

d. 5 to 5.3

Solution. (c) 4.4. Reference – Read the text below Sol: Multiplying factors of different bones - Femur – 3.6 to 3.8

- Tibia – 4.2 to 4.5 - Fibula – 4.4 - Humerus – 5 to 5.3 - Radius – 6.3 to 6.9 - Ulna – 6 to 6.3

Correct Answer. c

(188). An 18-year-old male surfer presents with a jellyfish sting. All the following are acceptable for rinsing and will not cause nematocyst firing, except:

a. Acetic acid solution

b. Isopropyl alcohol

c. Sterile water

d. Baking soda solution

Solution. (c) Sterile water Ref: Read the text below. Sol. - All the solutions mentioned may be used to rinse the nematocyst without causing firing, except for sterile water. - The hypotonic solution will trigger the nematocyst to fire. - The nematocyst can then be removed safely with forceps.

Correct Answer. c

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 80/86 (189). Within how much minutes of death the blood in most corpses, dead from natural or non- natural causes, becomes permanently incoagulable

a. 15-30 minutes

b. 30-60 minutes

c. 60-90 minutes

d. 90-120 minutes

Solution. (b) 30-60 minutes Ref: Read the text below. Sol. - Within about 30-60 minutes of death the blood in most corpses, dead from natural or non-natural causes, becomes permanently incoagulable. - This is due to the release of fibrinolysins, especially from small calibre vessels, e.g. capillaries, and from

serous surfaces, e.g. the pleura. Clots may persist when the mass of clot is too large to be liquified by the fibrinolysin available at the site of clot formation. - In some deaths associated with infection and cachexia, this fibrinolytic effect may fail to develop, explaining the presence of abundant clot in the heart and large calibre vessels. - Thus, in cases of sudden death the blood remains spontaneously coagulable only during a brief period immediately following death; it then becomes completely free from fibrinogen and will never again clot.

Correct Answer. b

(190). Which of the following amino acids is glycogenic only ?

a. Aspartate

b. Isoleucine

c. Leucine

d. Lysine

Solution. (a) Aspartate Ref:Read the text below Sol: - Aspartate is converted to oxaloacetate, which is glycogenic. - Leucine is ketogenic only. Isoleucine, lysine, and tyrosine are both glycogenic and ketogenic. In whole animal experiments, lysine is metabolized to both glucose and ketone bodies. - The reactions responsible for the conversion of lysine to carbohydrate in humans is unknown.

Correct Answer. a

(191). The metabolism of which of the following amino acids leads to the production of small amounts of nicotinic acid in humans ?

a. Cysteine

b. Methionine

c. Serine

d. Tryptophan

Solution. (d) Tryptophan Ref:Read the text below Sol: - Tryptophan is the only amino acid that is metabolized to form a vitamin in humans. The amount of nicotinic acid formed, however, is minor, and niacin is an essential human dietary constituent.

Correct Answer. d

Copyright © 2014 Delhi Academy of Medical Sciences, All Rights Reserved. 81/86 (192). The biosynthesis of which of the following amino acids is associated with the splitting of ATP to form ADP and Pi ?

a. Glutamine

b. Asparagine

c. Tyrosine

d. Glycine

Solution. (a) Glutamine Ref:Read the text below Sol: - Glutamine synthetase catalyzes a reaction between glutamate, ammonia, and ATP to form glutamine, ADP, and Pi, Asparagine synthetase catalyzes a reaction between asparate, ATP, and glutamine to yield AMP, PPi, glutamate, and asparagine.

Correct Answer. a

(193). Regarding hemoglobin, all are true except:

a. Deoxyhemoglobin is the most important buffer of human body

b. One molecule of Hemoglobin binds with four molecules of oxygen on complete oxygenation

c. 2,3 DPG favours ‘R’ state of hemoglobin.

d. Irrespective of the degree of deoxygenation, only one molecule of 2,3 DPG is bound to hemoglobin.

Solution. (c) 2,3 DPG favours ‘R’ state of hemoglobin Ref:Read the text below Sol: - 2,3 DPG is highly anionic. It is lies in a central cavity made of basic amino acids- positively charged.

- This creates a huge amount of electrostatic attraction and hence the Hb molecule is stabilized in T or taut structure,hence 2,3 DPG favours deoxygenation. - ince it is present in a central cavity made by all the four globin chains, only one molecule of 2,3 DPG is present irrespaective of the state of deoxygenation.

Correct Answer. c

(194). Which of the following enzymes participates in DNA repair in humans ?

a. DNA polymerase α

b. DNA polymerase β

c. DNA polymerase γ

d. DNA polymerase sigma

Solution. (b) DNA polymerase β Ref:Read the text below Sol: - DNA polymerase β is responsible for repair synthesis, and polymerase γ is responsible for replication of the mitochondrial genome. - Polymerase α is responsible for lagging strand synthesis, and polymerase ζ is responsible for leading strand biosynthesis

Correct Answer. b

a. 5' to 3' exonuclease activity

b. 3' to 5' exonuclease activity

c. RNase activity

d. Topoisomerase activity

Solution. (b) 3' to 5' exonuclease activity Ref:Read the text below Sol: - 3' to 5' exonuclease activity removes the last deoxynucleotide incorporated into the growing primer strand.

- If a noncomplementary base is added to the primer, the mismatched base is removed by proofreading activity. 5' to 3' exonuclease is used for the removal of oligonucleotide primers and for repair synthesis. - Topoisomerase activity adjusts the extent of supercoiling. - DNA ligase removes the nick between a free 3'-hydroxyl group and a 5'-phosphate group.

Correct Answer. b

(196). Which one of the following compounds is a key intermediate in the synthesis of both triacylglycerols and phospholipids?

a. CDP-choline

b. Phosphatidate

c. Triacylglyceride

d. Phosphatidylserine

Solution. (b) Phosphatidate Ref– Read the text below Sol: - Diacylglycerol-3-phosphate, more commonly known as phosphatidate, is an intermediate common to the synthesis of both triacylglycerol and phospholipids. - In a two-step process, glycerol phosphate is successively acylated by two acyl CoAs to lysophosphatidate, which contains a fatty acid group in the 1′position, and then phosphatidate, which contains fatty acid groups in the 1′

and 2′positions with a phosphate group in the 3′ position. - From that point, pathways for synthesis of phospholipids and triacylglycerol diverge. - If storage lipid is to be produced, phosphatidate is dephosphorylated by a phosphatase and then acylated by acyl CoA to form triacylglycerol. In contrast, if phospholipids are to be produced, phosphatidate is activated by CTP in a reaction that produces CDP- diacylglycerol and pyrophosphate. - Phosphatidylserine,phosphatidylinositol, phosphatidylethanolamine, and phosphatidylcholine can all be derived from CDP- diacylglycerol.

Correct Answer. b

a. Only in mitochondria of all mammalian tissues

b. Only in the cytosol of all mammalian tissues

c. Only in the endoplasmic reticulum of all mammalian tissues

d. In both the cytosol and mitochondria

Solution. (d) In both the cytosol and mitochondria Ref– Read the text below Sol: - The synthesis of 3-hydroxy-3- methylglutaryl CoA requires the condensation of three acetyl CoA groups. - The two enzymatic steps involved are the first two steps of cholesterol synthesis and ketone body synthesis. While cholesterol synthesis occurs in the cytosol of most mammalian tissues, ketone body synthesis can only occur in the mitochondria of liver cells. - Not only are cholesterol synthesis and ketone body synthesis separated by compartmentalization,they are separated by metabolic needs. - Cholesterol synthesis is an anabolic pathway that takes place when acetyl CoA production from excess dietary precursors is possible. - In contrast, ketone body production by the liver occurs when acetyl CoA levels from βoxidation are high. This catabolic situation exists during fasting, starvation, and uncontrolled diabetes.

Correct Answer. d

(198). Which of the following would rule out hyperuricemia in a patient?

a. Lesch-Nyhan syndrome

b. Xanthine oxidase hyperactivity

c. Carbamoyl phosphate synthase deficiency

d. Purine overproduction secondary to Von Gierke’s disease

Solution. (c) Carbamoyl phosphate synthase deficiency Ref– Read the text below Sol: - Carbamoyl phosphate (CAP) synthase I is found in mitochondrial matrix and is the first step in urea synthesis, condensing CO2 and NH4+ . Hyperammonemia occurs when CAP is deficient.

- CAP synthase II forms CAP as the first step in pyrimidine synthesis. Its complete deficiency would probably be a lethal mutation. - When its activity is decreased, purine catabolism to uric acid is decreased, decreasing the possibility of hyperuricemia. In contrast, gout, Lesch-Nyhan syndrome,high xanthine oxidase activity, and von Gierke’s disease [glycogen storage disease type Ia (232200)] all lead to increased urate production and excretion.

Correct Answer. c

a. Aspartate

b. Carbamoyl phosphate

c. Carbon dioxide

d. Glutamine

Solution. (a) Aspartate Ref– Read the text below Sol: - During purine ring biosynthesis, the amino acid glycine is completely incorporated to provide C4, C5, and N7. - Glutamine contributes N3 and N9, aspartate provides N1, and derivatives of tetrahydrofolate furnish C2 and C8. Carbon dioxide is the source of C6. - In pyrimidine ring synthesis, C2 and N3 are derived from carbamoyl phosphate, while N1, C4, C5, and C6 come from aspartate.

Correct Answer. a

a. 0.33

b. 0.67

c. 1.0

d. 1.33

Solution. (b) 0.67 Ref: Read the text below. Sol. - In India the average rate of this fall varies between 0.50C (in summer) to 0.70C or more in cold places. - There is however a “lag period” before post mortem cooling begins, because body temperature is maintained until that of the internal organs is dissipated by conduction and convection. - This lag period (30 min to 2 hours) shouldbe added to the time passed since death.

Correct Answer. b

Test Answer 1.(c) 2.(a) 3.(c) 4.(d) 5.(a) 6.(b) 7.(a) 8.(d) 9.(a) 10.(a)

11.(c) 12.(a) 13.(c) 14.(c) 15.(a) 16.(d) 17.(b) 18.(c) 19.(b) 20.(d)

21.(b) 22.(a) 23.(c) 24.(b) 25.(a) 26.(b) 27.(d) 28.(d) 29.(b) 30.(c)

31.(a) 32.(d) 33.(c) 34.(b) 35.(d) 36.(d) 37.(d) 38.(c) 39.(d) 40.(a)

41.(d) 42.(a) 43.(c) 44.(b) 45.(c) 46.(a) 47.(a) 48.(a) 49.(a) 50.(b)

51.(d) 52.(d) 53.(d) 54.(a) 55.(c) 56.(b) 57.(a) 58.(d) 59.(c) 60.(d)

61.(a) 62.(b) 63.(c) 64.(d) 65.(d) 66.(d) 67.(a) 68.(b) 69.(b) 70.(c)

71.(d) 72.(c) 73.(d) 74.(d) 75.(a) 76.(b) 77.(d) 78.(b) 79.(a) 80.(b)

81.(c) 82.(b) 83.(c) 84.(a) 85.(d) 86.(b) 87.(b) 88.(d) 89.(d) 90.(d)

91.(b) 92.(b) 93.(a) 94.(d) 95.(b) 96.(b) 97.(d) 98.(a) 99.(a) 100.(b)

101.(c) 102.(a) 103.(a) 104.(a) 105.(a) 106.(c) 107.(a) 108.(a) 109.(a) 110.(b)

111.(a) 112.(c) 113.(a) 114.(a) 115.(b) 116.(a) 117.(a) 118.(b) 119.(c) 120.(b)

121.(c) 122.(c) 123.(a) 124.(a) 125.(d) 126.(a) 127.(a) 128.(d) 129.(b) 130.(d)

131.(b) 132.(d) 133.(c) 134.(a) 135.(c) 136.(c) 137.(d) 138.(b) 139.(c) 140.(b)

141.(a) 142.(b) 143.(a) 144.(c) 145.(a) 146.(a) 147.(d) 148.(b) 149.(b) 150.(c)

151.(b) 152.(b) 153.(d) 154.(c) 155.(d) 156.(a) 157.(a) 158.(d) 159.(b) 160.(c)

161.(a) 162.(b) 163.(d) 164.(d) 165.(c) 166.(c) 167.(b) 168.(b) 169.(d) 170.(c)

171.(d) 172.(b) 173.(b) 174.(d) 175.(d) 176.(c) 177.(b) 178.(c) 179.(a) 180.(c)

181.(a) 182.(d) 183.(a) 184.(b) 185.(b) 186.(c) 187.(c) 188.(c) 189.(b) 190.(a)

191.(d) 192.(a) 193.(c) 194.(b) 195.(b) 196.(b) 197.(d) 198.(c) 199.(a) 200.(b)