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Test Information
Test Name
Biochemistry & FMT (SWT MD/MS)
Total Questions
200
Test Type
Examination
Difficulty Level
Difficult
Total Marks
600
Duration
120minutes
Test Question Language:- ENGLISH
(1).
Pyruvate can be converted directly to all of the following except?
a. Phosphoenol pyruvate
b. Alanine
c. Acetyl Co A
d. Lactate
Solution. Ans-1 : (a) Phosphoenol pyruvate
Ref.:Read the text below
Sol :
For the conversion of pyruvate to phosphoenolpyruvate, pyruvate has to undergo carboxylation and decarboxylation reactions by
pyruvate carboxylase and phosphoenol pyruvate carboxykinase respectively..
Pyruvate may get converted to alanine by transamination reaction.
Pyruvate may get converted to acetyl Co A by Pyruvate Dehydrogenase Complex (PDH complex)
Pyruvate may get converted to lactate by LDH enzyme.
Correct Answer. a
(2).
Insulin causes lipogenesis by all except?
a. Increasing acetyl Co A carboxylase activity
b. Increases the transport of glucose in to the cell
c. Inhibits PDH
d. Decreases intracellular cAMP level
Solution. Ans-2: (c) Inhibits PDH
Ref.:Read the text below
Sol :
Insulin stimulates activity of PDH complex and thus enhances oxidative decarboxylation of pyruvate converting it to acetyl Co A.
During lipogenesis insulin stimulates acetyl Co A carboxylase activity and thus stimulates lipogenesis.
Insulin increases transportation of glucose in to the cell by enhancing the recruitment of GLUT to the cell membrane.
Insulin decreases level of cAMP in the cell by inhibiting the activity of adenylyl cyclase enzyme.
Correct Answer. c
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(3).
All are features of a normally decomposed body except :
a. Greenish discoloration of skin over caecum
b. Blisters over skin
c. Mummification
d. Small military skin granules
Solution. Ans-3 : (c) Mummification
Ref.:Read the text below
Sol :
·
A general description of postmortem changes due to decomposition basically includes two stages of autolysis, and four stages of
putrefaction, besides some conservative phenomena such as saponification or adipocere, natural mummification, calcification, etc.
·
However, these latter events only occur in specific conditions. Autolysis consists of the fast and intense spontaneous self-destruction
of tissues by the body enzymes present in the cells, without any bacterial interference.
·
Once cells stop receiving nutrients and oxygen via blood circulation, they start anaerobic (without oxygen) "breathing", breaking
ATP (adenosine tri-phosphate) into ADP (adenosine di-phosphate) to obtain energy.
·
Anaerobic respiration lasts for a few hours, until all ATP reserves are exhausted. The anaerobic respiration induces the accumulation
of lactic acid in cell tissues that disrupts cell function. Enzymes then collapse the cell nucleus and cell breakdown (necrosis) occurs.
Correct Answer. c
(4).
All are true about gunshot entry wound except
a. Entry would is smaller than exit wound
b. Entry would is surrounded with cornea
c. Entry would is everted
d. Entry would is inverted
Solution. Ans-4: (c) Entry would is everted
Ref.:Read the text below
Sol :
·
Entrance Wounds
·
The entrance wound is normally smaller and quite symmetrical in comparison to the exit wound, which can sometimes be ragged
with skin, tissue, and muscle and bone damage. Entrance wounds are often ringed with the residue of gunpowder and cordite - the two
substances contained within a bullet.
·
A close range gunshot - if the weapon is touching the victim's body - will normally have what is known as an 'abrasion ring' and also
a clear imprint of the weapon's barrel. This can go some way to identifying the weapon the round was fired from, which is useful as in
most cases the victim will not have any predisposition to the nature of firearms and identifying it will prove difficult if they survive the
gunshot wound.
·
Entrance wounds will usually have some kind of discolouration around them; perhaps a black or grey ring caused by the burning of
the gunpowder as it makes contact with the skin.
·
'Tattooing' as it is sometimes referred to, is when the gunpowder will spray around the area of the wound and burn to the skin as it
is hot on being fired from a weapon. The same process can be applied to rounds fired from shotguns were the rounds are filled with
pellets or 'buckshot'.
·
Exit Wounds
·
Exit wounds - as we have already mentioned - are usually larger than the entrance wound and this is because as the round moves
through the body of the victim it slows down and explodes within the tissue and surrounding muscle. This slowing down of the projectile
means that as it reaches the end of its trajectory it has to force harder to push through. This equates to the exit wound normally looking
larger and considerably more destructive than its pre-cursor - the entrance wound.
·
Exit wounds will often bleed profusely as they are larger but entrance wounds can sometimes look only like small holes - unless the
weapon is fired at close proximity to the victim.
Correct Answer. c
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(5).
Features of strychnine poisoning :
a. Gradual onset of symptoms
b. Duration of rigor mortis is longer
c. All muscles not affected at given time
d. Between fits muscles are slightly rigid
Solution. Ans-5 : (b) Duration of rigor mortis is longer
Ref.:Read the text below
Sol :
·
The clinical signs of strychnine poisoning relate to its effects on the central nervous system.
·
The first clinical signs include uneasiness, restlessness, anxiety, muscle twitching and stiffness of the neck.
·
These signs can resemble tetanus, withan increase in spinal reflexes leading to tonic convulsions characterized by sudden
contractions of all striated muscles followed by complete relaxation.
·
Rigor mortis occurs shortly after death and persists for days
Correct Answer. b
(6).
Cheiloscopy is the study of :
a. Palato prints
b. Nail prints
c. Foot prints
d. Lip prints
Solution. Ans-6: (d) Lip prints
Ref.:Read the text below
Sol :
·
Cheiloscopy is a forensicinvestigation technique that deals with identification of humans based on lips traces.
·
The aim of this study is to establish the uniqueness of lip prints which aids in personal identification.
Correct Answer. d
(7).
True statement about glucokinase is?
a. Km value is higher than normal blood sugar
b. Found exclusively in liver
c. G-6 P inhibit it
d. Has both G-6-phosphatase and pyruvate kinase activity
Solution. Ans-7: (a) Km value is higher than normal blood sugar
Ref.:Read the text below
Sol :
Glucokinase is an an isoenzyme of hexokinase and is found in liver and cell of the pancreas.
It has got low affinity (High km) for glucose and so is able to remove glucose from blood only when glucose is more than 100 mg/d.
Glucose 6 phosphate has got no effect on action of glucokinase.
Glucose 6 phosphase has got inhibitory effect on hexokinase activity.
Correct Answer. a
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(8).
In glycolysis, the first committed step is catalysed by?
a. 2, 3 BPG kinase
b. Glucokinase
c. Hexokinase
d. Phosphofructokinase
Solution. Ans-8: (d) Phosphofructokinase
Ref.:Read the text below
Sol :
In glycolysis, PFK-1 converts fructose-6-phosphate to fructose—6-bisphosphate, which is destined only for glycolysis and no other
pathway.
Action of hexokinase and glucokinase converts glucose to glucose-6-phosphate.
Glucose 6 phosphate has got many additional fates also, in addition of entering in glycolysis. So, the step catalysed by
phosphofructokinase is the committed step of glycolysis.
Correct Answer. d
(9).
Regarding synthesis of triacylglycerol in the adipose tissue, all of the following are true except?
a. Synthesis from dihydroxyacetone phosphate
b. Glycerol kinase enzyme plays an important role
c. Glycerol 3 phosphate dehydrogenase plays an important role
d. Phosphatidate is hydrolysed
Solution. Ans-9 : (b) Glycerol kinase enzyme plays an important role
Ref.:Read the text below
Sol :
Adipose tissue does not possess glycerol kinase enzyme, so for the synthesis of triacylglyserol, adipose tissue is dependent on glycolysis
which provides dihydroxyacetone phosphate which gets converted to Glycerol 3 phosphate with the help of enzyme glycerol 3 phosphate
dehydrogenase which catalyses a reversible reaction.
Now this glycerol 3 phosphate acts as a nucleus on which fatty acid gets esterified to form triacylglyserol.
Correct Answer. b
(10).
A patient diagnosed with isolated increase of LDL. His father and brother had the same disease with increased cholesterol. The likely
diagnosis is?
a. Familial type III hyperlipoproteinemia
b. Abetalipoproteinemia
c. Familial lipoprotein lipase deficiency
d. LDL receptor mutation
Solution. Ans-10 : (d) LDL receptor mutation
Ref.:Read the text below
Sol :
LDL receptor mutation is seen in Type IIa hyperlipoproteinemia.
Correct Answer. d
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(11).
Which of the following facts regarding a scar is not correct ?
a. An abrasion does not result in scar.
b. A scar remains vascular up to 6 months
c. A scar is devoid of elastic tissue
d. A scar is devoid of sweat glands
Solution. Ans-11 : (b) A scar remains vascular up to 6 months
Ref.:Read the text below
Sol :
Scarsare areas of fibrous tissue (fibrosis) that replace normal skinafter injury. A scar results from the biological process of woundrepair
in the skin and other tissuesof the body. Thus, scarring is a natural part of the healingprocess
·
An abrasion does not result in scar.
·
A scar is devoid of elastic tissue
·
A scar is devoid of sweat glands
Correct Answer. b
(12).
To carry out a medicolegal autopsy, which one of the following is not an essential pre-requisite ?
a. The consent of the relative
b. The complete inquest report
c. Identification by the police officer accompanying the body
d. Identification by a relative or two accompanying the body
Solution. Ans-12 : (a) The consent of the relative
Ref.:Read the text below
Sol :
Medicolegal autopsy is conducted to solve a specific question of law, and the consent of relatives is of no consequence.
However in a different kind of autopsy – the hospital autopsy, or the pathological autopsy – which is done merely for research, the
consent of relatives is a must.
Correct Answer. a
(13).
When a police officer or magistrate submits the dead body for postmortem to a doctor, generally it is implied that the autopsy would be
done on :
a. The whole body
b. Only those parts which are injured
c. Only those parts which are not injured
d. Only those parts for which relatives give permission
Solution. Ans-13 : (a) The whole body
Ref.:Read the text below
Sol :
By and large medicolegal autopsies are done on the whole body. In hospital autopsies, where autopsies are conducted only for academic
reasons, limited autopsies may have to be done if relatives do not give consent for complete autopsies.
Correct Answer. a
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(14).
After fixation in formalin, the weight of the brain :
a. Remains unchanged
b. Increases by about 5-10%
c. Decreased by about 5-10%
d. Increases in cases of males, and decreases in cases of females
Solution. Ans-14 : (b) Increases by about 5-10%
Ref.:Read the text below
Sol :
·
After fixation in formalin, the weight of the brain increases by about 5-10%
Correct Answer. b
(15).
Lower end of femur can help to determine the :
a. Age and sex
b. Height and sex
c. Weight and sex
d. Only age
Solution. Ans-15 : (a) Age and sex
Ref.:Read the text below
Sol :
§ It is common knowledge that lower end of femur can help in age determination (e.g. by seeing whether the epiphysis has appeared or
not, and whether it has united with the shaft or not).
§ What is not commonly known is that it can be used to determine sex also. In fact four different features of femur can be used to
determine sex.
§ These are (i) Vertical diameter of the head (ii) Popliteal length (iii) Bicondylar width and (iv) Trochanteric oblique length
Major differences between female and a male femur
Criteria
Female
Male
Vertical diameter of head
< 41.5 mm > 45.5 mm
Popliteal length
< 106 mm > 145 mm
Bicondylar width
< 72 mm
> 78 mm
Trochanteric oblique length < 390 mm > 450 mm
Correct Answer. a
(16).
In a badly decomposed body, all of the following features would help in its identification except ?
a. Bones
b. Color of the iris
c. Hair
d. Teeth
Solution. Ans-16 : (b) Color of the iris
Ref.:Read the text below
Sol :
Color of the iris is not very helpfulin identification in any case, least of all in decomposed bodies.
Hair, their length, and their color, especially if they were dyed are helpful in identification.
Bones can help in identification of sex, stature etc. teeth are unique in each individual and are extremely valuable in identification.
Correct Answer. b
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(17).
What is the time limit of exhumation in India ?
a. 10 years
b. 30 years
c. 35 years
d. No time limit
Solution. Ans-17 : (d) No time limit
Ref.:Read the text below
Sol :
·
In India, there is no time limit for ordering of the exhumation, but many western countries have well-defined time limit up to which
exhumation can be done.
·
For instance, in France the time limit is 10 years, and in Germany the time limit is 30 years. Thus in France, if in the 11th year (say)
of death, some relevant facts are found which reveal foul play, even then the body can not be exhumed.
Correct Answer. d
(18).
The best time for exhumation is :
a. Early morning
b. Late in the evening
c. Early night
d. Midnight
Solution. Ans-18 : (a) Early morning
Ref.:Read the text below
Sol :
§ Although law does not specify any time for exhumation, itis preferable to conduct exhumation early morning, so the whole process of
digging, and autopsy if required, can be completed during the day.
Correct Answer. a
(19).
Mummification is likely to occur in which of the following environment?
a. Cold and dry
b. Warm and dry, with hot air blowing constantly
c. Cold and humid, with icy cold winds blowing
d. Warm and humid, with humid air blowing
Solution. Ans-19 : (b) Warm and dry, with hot air blowing constantly
Ref.:Read the text below
Sol :
·
The natural mummification process usually happens in extremely dry environments that allow the fast dehydration of tissues,
simultaneously slowing down or inhibiting the decomposition by bacteria and other microorganisms.
·
Mummification is likely to occur in Warm and dry, with hot air blowing constantly
Correct Answer. b
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(20).
Which of the following facts regarding treatment of poisoning is not correct ?
a. The most useful and harmless household emetic is a saturated solution of common salt
b. Ipecac Fluid extract should never be used for inducing emesis
c. Syrup of Ipecac is the most useful household emetic
d. Apomorphine as an emetic is not effective orally
Solution. Ans-20 : (a) The most useful and harmless household emetic is a saturated solution of common salt
Ref.:Read the text below
Sol :
Saturated solution of common salt can cause hypernatremia and is considered very dangerous to use as an emetic.
There is a difference between Ipecac Fluid Extract and syrup of Ipecac.
Ipecac Fluid extract is 14 times more potent and may cause fatalities. It (Ipecac Fluid extract) is never used as an emetic.
Apomorphine is not effective orally and must be given parenterally, usually by the subcutaneous route.
Correct Answer. a
(21).
Which of the following statement is not true ?
a. Apomorphine given as an emetic is not effective orally
b. Syrup of Ipecac may be effective as an emetic even in antiemetic poisoning such as that by phenothiazines
c. Syrup of Ipecac should never be followed by a drink of water, as it may cause fulminant vomiting causing rupture of oesophagus
d. Ipecac can produce toxic effects on the heart because of its content of emetine
Solution. Ans-21 : (c) Syrup of Ipecac should never be followed by a drink of water, as it may cause fulminant vomiting causing rupture
of oesophagus
Ref.:Read the text below
Sol :
Apomorphine is indeednot effective orally. It has to be given parenterally, usually by the subcutaneous route.
Syrup of Ipecac acts both on CTZ, as well as on gastric mucosa, so it may be useful in antiemetic poisoning.
Syrup of ipecac is usually followed by a drink of water because emesis may not occur when the stomach is empty.
Correct Answer. c
(22).
Which of the following facts about gastric lavage are true ?
a. Gastric lavage must be undertaken irrespective of the fact whether the patient has ingested a toxic or a non-toxic agent
b. Endotracheal intubation must be done prior to gastric lavage in a comatosed patient
c. The length of the tube to be inserted is 100 cm
d. Gastric lavage must be performed even if the patient does not give consent
Solution. Ans-22 : (b) Endotracheal intubation must be done prior to gastric lavage in a comatosed patient
Ref.:Read the text below
Sol :
Since gastric lavage is essentiallya traumatic procedure, it must not be used when the patient has ingested a non-toxic agent or a nontoxic dose of a poisonous agent.
The length of the tube to be inserted is 50 cm in adults and 25 cm for a child.
Performing gastric lavage without consent may amount to assault u/s 351 of I.P.C. the preferred route of insertion of gastric lavage tube
is oral and nasal.
Correct Answer. b
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(23).
Cholesterol present in LDL?
a. Binds to a cell receptor and diffuses across the cell membrane
b. When it enters the cell it suppress the activity of ACAT
c. Once in the cell is converted to cholesteryl ester by LCAT
d. Once it has accumulated in the cell, inhibits replenishment of LDL receptor.
Solution. Ans-23 : (d) Once it has accumulated in the cell, inhibits replenishment of LDL receptor.
Ref.:Read the text below
Sol :
Cholesterol doesnot diffuse within the cell rather the whole LDL particle is engulfed via cholesterol receptor.
This cholesterol of pool activates ACAT function and has no effect on LCAT function.
It is converted to cholesterol ester by activity of ACAT (not LCAT). Cholesterol of the pool has got inverse effect on receptor synthesis.
Correct Answer. d
(24).
Acetyl Co A acts as a substrate for all the enzymes except?
a. HMG Co A synthetase
b. Malic enzymes
c. Malonyl Co A synthetase
d. Fatty acid synthetase
Solution. Ans-24 : (b) Malic enzymes
Ref.:Read the text below
Sol :
HMG Co A synthetaserequire acetyl Co A as a precursor molecule. HMG Co A in turn is responsible for cholesterol synthesis and ketone
body synthesis.
Malonyl Co A synthetaseacts on acetyl Co A and carboxylates it to form malonyl Co. A.
Fatty acid synthetasecomplex acts on acetyl Co A and malonyl Co A as precursor molecule for fatty acid synthesis.
Malic enzymedoes not require acetyl Co A as the precursor molecule.
Malic enzymeis NADP+ dependent malate dehydrogenase which converts malate to pyruvate.
Correct Answer. b
(25).
Ketone body which is maximum in diabetic ketoacidosis?
a. Acetone
b. Pyruvate
c. Acetoacetic acid
d. β-(OH) butyrate
Solution. Ans-25 : (d) β-(OH) butyrate
Ref.:Read the text below
Sol :
Acetoacetic acidis the precursor ketone body and b–(OH) butyrate which is formed from acetoacetate is most abundant ketone body in
the circulation.
Conversion of acetoacetate to b–(OH) butyrate require presence of b–(OH) butyrate dehydrogenase which requires NAD as a coenzyme.
Correct Answer. d
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(26).
Which of the following is an activator of LCAT?
a. Apo B-100
b. Apo B-48
c. Apo A IV
d. Apo A I
Solution. Ans-26 : (d) Apo A I
Ref.:Read the text below
Sol :
Lecithin cholesterol acyl transferase (LCAT) is the enzyme which is found on the surface of HDL and it transfers acyl moiety from a
phospholipids (lecithin) to the cholesterol and in this process lecithin is converted to lysolecithin, and cholesterol is converted to
cholesterol ester. Apo A1 acts as activating cofactor for LCAT enzyme.
Correct Answer. d
(27).
If choline moiety is replaced by ethanolamine, the net product is?
a. Cerebroside
b. Sphingomyelin
c. Cephalin
d. Plasminogen
Solution. Ans-27 : (c) Cephalin
Ref.:Read the text below
Sol :
Phosphatidylethanolamine is also known as cephalin.
Phsophatidylcholine isalso known as lecithine.
Correct Answer. c
(28).
Apolipoprotein A is found in?
a. HDL
b. LDL
c. Chylomicron
d. VLDL
Solution. Ans-28 : (a) HDL
Ref.:Read the text below
Sol :
Apo A is importantcomponent of HDL. Apo A is activator of LCAT which is important enzyme for reverse cholesterol transport,
converting cholesterol to cholesterol ester.
Correct Answer. a
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(29).
Ligand for LDL particle is
a. APO-B 100
b. APO – E
c. APO 48
d. APO A11
Solution. Ans-29: (a) APO-B 100
Ref.:Read the text below
Sol :
LDL has got only one apolipoproteinon it and it is APO-B 100. It acts as ligand as it is being recognized by LDL receptor.
Correct Answer. a
(30).
Lack of α–oxidation of fatty acid leads to ?
a. Accumulation of phytanic acid
b. Formation of dicarboxylic acid
c. Formation of proprionic acid
d. Oxidation of branched chain fatty acid
Solution. Ans-30 : (a) Accumulation of phytanic acid
Ref.:Read the text below
Sol :
Phytanic acidis branched fatty acid found in dairy products and meat.
Impaired a–oxidation prevents b–oxidation and causes accumulation of fatty acid.
Correct Answer. a
(31).
Which of the following is the best antidote for oxalic acid poisoning?
a. BAL
b. Animal charcoal
c. Calcium gluconate
d. Magnesium sulphate
Solution. Ans-31 : (c) Calcium gluconate
Ref.:Read the text below
Sol :
Calcium gluconate combines with oxalic acid to form insoluble oxalates and thus prevents its absorption.
Both calcium gluconate and calcium lactate are effective.
The dose is 150 mg/kg orally.
Electrocardiogram with serum calciummonitoring are also required.
Renal failure (due to deposition of calcium oxalate crystals in proximal tubules) may require hemodialysis.
Correct Answer. c
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(32).
Which one among the following is not an antioxidant?
a. Selenium
b. Copper
c. Zinc
d. Iron
Solution. Ans-32 : (d) Iron
Ref.:Read the text below
Sol :
Iron is not an antioxidant.
In can catalyse the formation of reactive oxygen species and cause oxidant damage.
Selenium, copper, zinc are examples of antioxidants
Antioxidants:
- Vitamins: C, E and A (beta carotene) (Think: “ACE”)
- Minerals: Selenium, Zinc, Copper (minor)
- Others : Cysteine, Glutathione, Caffeine, allopurinol
Transport proteins: transferring, ferritin, lactoferrin, ceruloplasmin.
Correct Answer. d
(33).
Glucose-galactose malabsorption is related with:
a. SLC5A1 gene
b. SLC5A2 gene
c. SLC5A3 gene
d. SLC5A4 gene
Solution. Ans-33 : (a) SLC5A1 gene
Ref.:Read the text below
Sol :
·
The SLC5A1 gene provides instructions for producing a sodium/glucose cotransporter protein called SGLT1.
·
This protein is found mainly in the intestinal tract and, to a lesser extent, in the kidneys, where it is involved in transporting glucose
and the structurally similar galactose across cell membranes.
·
The sodium/glucose cotransporter protein is important in the functioning of intestinal epithelial cells, which are cells that line the
walls of the intestine. These cells have fingerlike projections called microvilli that absorb nutrients from food as it passes through the
intestine. Based on their appearance, groups of these microvilli are known collectively as the brush border.
·
The sodium/glucose cotransporter protein is involved in the process of glucose uptake in the instesinal cells due to a sodium gradient
across the membrane.
·
This is a secondary active transport because the sodium gradient generated for the functioning of the sodium/calcium exchanger is
created by the sodium/potassium pump which requires ATP. Sodium and water are transported across the brush border along with the
sugars in this process.
Correct Answer. a
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(34).
Which of the following is an Omega 6 fatty acid?
a. Alpha linolenic acid
b. Linoleic acid
c. EPA
d. DHA
Solution. Ans-34 : (b) Linoleic acid
Ref.: Read the text below
Sol :
Essential Fatty Acids
Linoleic (most essential)
Alpha linolenic acid
Omega 3 Fatty Acids
Alpha linolenic acid
DHA (cervonic acid)
EPA (timodonic acid
Omega 6 Fatty Acids
Gamma linolenic acid
Arachidonic acid
Linoteic acid
Correct Answer. b
(35).
During prolonged starvation, the rate of gluconeogenesis depends on
a. Increased alanine levels in liver
b. Decreased cGMP in liver
c. ADP in liver
d. Decreased essential fatty acids in liver
Solution. Ans-35 : (a) Increased alanine levels in liver
Ref: Read the text below
Sol :
In the fasting state, there is a considerable output of alanine from skeletal muscle, far in excess of its concentration in the muscle
proteins that are being catabolized.
It is formed by transamination of pyruvate produced by glycolysis of muscle glycogen, and is exported to the liver.
In the liver after transamination alanine is converted back to pyruvate, which is a substrate for gluconeoenesis.
Correct Answer. a
(36).
All of the following statement about lipoprotein lipase is true, except?
a. Found in adipose tissue
b. Found in myocytes
c. Deficiency leads to hypertriacylglyserolemia
d. Does not require CII as a cofactor
Solution. Ans-36 : (d) Does not require CII as a cofactor
Ref: Read the text below
Sol :
Lipoprotein lipase is extracellular enzyme adhered to the endothelial cell of blood vessels supplying various organs.
Heparan sulphate is required to adhere this enzyme to the endothelial cell.
High density of negative charges in the heparin sulphate bring the positive charged molecule of lipoprotein lipase in the vicinity and
holds them by electrostatic interaction.
Lipoprotein lipase activity is stimulated by Apolipoprotein CI and CII.
So triacyglycerol of VLDL and CM is acted upon by this enzyme and fatty acid is send to the organ.
Correct Answer. d
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(37).
Regarding synthesis of triacylglycerol in the adipose tissue, all of the following are true except?
a. Synthesis from dihydroxyacetone phosphate
b. Glycerol kinase enzyme plays an important role
c. Glycerol 3 phosphate dehydrogenase plays an important role
d. Phosphatidate is hydrolysed
Solution. Ans-37 : (b) Glycerol kinase enzyme plays an important role
Ref: Read the text below
Sol :
Adipose tissue does not possess glycerol kinase enzyme, so for the synthesis of triacylglyserol, adipose tissue is dependent on glycolysis
which provides dihydroxyacetone phosphate which gets converted to Glycerol 3 phosphate with the help of enzyme Glycerol 3 phosphate
dehydrogenase which catelyses a reversible reaction.
Now this glycerol 3 phosphate acts as a nucleus on which fatty acid gets esterified to form triacylglyserol.
Correct Answer. b
(38).
Carboluria is seen in acute poisoning with :
a. Nitric acid
b. Sulphuric acid
c. Oxalic acid
d. Phenol
Solution. Ans-38 : (d) Phenol
Ref.:Read the text below
Sol :
Carboluria is the presence of phenol (also knownas carbolic acid) and its metabolic products, catechol (pyrocate-chol or 1,2Dihydroxybenzene) and quinol (hydroxyquinone or 1, 4 – dihydroxybenzene) in urine.
Correct Answer. d
(39).
Garlic like odor is present in breath and excreta in poisoning by :
a. Cyanide
b. Lead
c. Copper
d. Phosphorus
Solution. Ans-39 : (d) Phosphorus
Ref.:Read the text below
Sol :
Poisons which cause a garlic like smell in breath are arsenic, dimethyl sulfoxide, phosphorus, selenium and tellurium.
Correct Answer. d
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(40).
The usual fatal period from poisoning by yellow phosphorus is about :
a. 6 – 7 minutes
b. 6 – 7 hours
c. 6 – 7 days
d. 6 – 7 weeks
Solution. Ans-40 : (c) 6 – 7 days
Ref.:Read the text below
Sol :
The usual fatal period in phosphorus poisoning is 6 – 7 days. However if there is very fulminant poisoning, the victim may die from
collapse within 24 hours.
Since in this question the usual fatal period is asked, choice (c) is the correct choice.
Correct Answer. c
(41).
Most specific test for organophosphorus poisoning is :
a. RBC cholinesterase level
b. Plasma cholinesterase level
c. RBC uroporphyrin level
d. Measurement of serum levels of organophosphorus
Solution. Ans-41 : (a) RBC cholinesterase level
Ref.:Read the text below
Sol :
RBC cholinesterase levels are better indicators of acute or chronic organophosphorus poisoning than plasma cholinesterase levels,
because RBC cholinesterase reflects the true cholinesterase levels.
Plasma cholinesterase levels represent the pseudocholinesterase (which is a protein synthesized in liver) levels.
Correct Answer. a
(42).
When ingested, endrin is stored mostly in :
a. Brain
b. Liver
c. Bones
d. Fatty tissues
Solution. Ans-42 : (d) Fatty tissues
Ref.:Read the text below
Sol :
Because of this peculiar fact, in poisoning with endrin, samples of body fats (and blood) are taken for analysis.
Correct Answer. d
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(43).
Lathyrus sativus or kesari dal contains a :
a. Neurotoxin
b. Cardiotoxin
c. Hemolytic toxin
d. Myotoxin
Solution. Ans-43: (a) Neurotoxin
Ref.:Read the text below
Sol :
·
Like other grain legumes, L. sativus produces a high-proteinseed. The seeds also contain variable amounts of a neurotoxicamino
acidβ-N-Oxalyl-L-α,β-diaminopropionic acidor ODAP or BOAA
·
ODAP is considered as the cause of the disease neurolathyrism, a neurodegenerativedisease that causes paralysisof the lower body:
emaciation of Gluteal muscle (buttocks). Thedisease has been seen to occur after faminesin Europe (France, Spain, Germany), North
Africa, South Asia, and is still prevalent in Eritrea, Ethiopiaand Afghanistan (pan handle) when Lathyrus seed is the exclusive or main
source of nutrients for extended periods.
·
Research has shown that ODAP concentration increasesin plants grown under stressful conditions, compounding the problem.
Correct Answer. a
(44).
Beta-N-oxalylamino – L – alanine (BOAA) is the toxic principle of :
a. Cassava
b. Mexican poppy
c. Fava beans
d. Lathyrus sativus
Solution. Ans-44 : (d) Lathyrus sativus
Ref.:Read the text below
Sol :
Beta-N-oxaly lamino-L-alanine (BOAA), is alsoreferred to as Beta N-oxaly-L-alpha-beta-diamino-propionic acid (ODAP).
It is a neurotoxin which mimics glutamate. This mimicry causes excessive triggering of the fast excitatory signals running through the
brain and nervous system resulting in exhaustion and death of neurons.
When consumed in large quantities, kesari dal causesa spastic disorder called lathyrism (sometimes called neurolathyrism.)
Correct Answer. d
(45).
A child has a total of 24 teeth. His probable age is :
a. Less than 6
b. Between 6 to 9 years
c. Between 6 to 11 years
d. Between 12 to 18 years
Solution. Ans-45 : (c) Between 6 to 11 years
Ref: Read the text below.
Sol:
The total number of teeth changes only when superadded teeth erupt. Molars are the only superadded teeth.
Molars erupt at 6, 12 and 18 years – a simple multiplication table to remember (the eruption of last is variable, but for the purposes of
this question we can take the age as 18.)
Thus the total number of teeth changes only at 6, 12 and 18.
Since one molar erupts in each quadrant at these ages, the total number of teeth will increase by 4.
Correct Answer. c
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(46).
Which of the following is not a criteria used in Gustafson’s method?
a. Root resorption
b. Periodontosis
c. Width of enamel prisms
d. Secondary dentin formation
Solution. Ans-46 : (c) Width of enamel prisms
Ref: Read the text below.
Sol:
Gustafson’s method uses the following
1. Attrition
2. Periodontitis
3. Secondary Dentine
4. Root Resorption
5. Cementum Apposition
6. Transparency of Root
Correct Answer. c
(47).
Putrefaction is delayed in :
a. Peritonitis
b. Septicemia
c. Anasarca
d. Zinc chloride poisoning
Solution. Ans-47: (d) Zinc chloride poisoning
Ref: Read the text below.
Sol:
Putrefaction
Rate
Delayed in
In air > water > earth
Carbolic acid
Zinc chloride
Heavy metals e.g. Arsenic, antimony
Strychnine (nux vomica)
Correct Answer. d
(48).
Ewing’s postulates deals with relationship between:
a. Trauma and tumor
b. Trauma and MI
c. Trauma and SAH
d. Trauma and poisoning
Solution. Ans-48: (a) Trauma and tumor
Ref: Read the text below.
Sol:
Ewing’s postulates deals with relationship between Trauma and tumor.
Ewing’s postulate should be satisfied before accepting a relationship between trauma and new growth.
Correct Answer. a
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(49).
Head of humerus unites with the shaft at the age of:
a. 12 years
b. 14 years
c. 16 years
d. 18 years
Solution. Ans-49 : (d) 18 years
Ref: Read the text below.
Sol:
Medial epicondyle unites with the shaft at 16 years.
Upper end of radius uniteswith the shaft at 16 years.
Lower end of radius uniteswith the shaft at 16 years. (These three centres fall upon the lowest line – ossification centres around elbow
joint).
Head of femur unites withthe shaft at 17 years. (This is the ossification center aroung hip joint – it falls on the second line from below)
Lower end of tibia uniteswith the shaft at 17 years. Lower end of fibula unites with the shaft at 17 years. (These two are the ossification
centers around ankle joint – it falls on the second line from below)
Head of humerus unites with the shaft at 18 years.
Correct Answer. d
(50).
When a group of muscles of a dead body were in state of strong compaction immediately prior to death and remain so even after death,
the condition is termed as:
a. Gas stiffening
b. Rigor mortis
c. Cadaveric spasm
d. Cold stiffening
Solution. Ans-50 : (c) Cadaveric spasm
Ref: Read the text below.
Sol:
Cadaveric Spasm (Instantaneous rigor/cataleptic rigidity
·
Rare condition caused by stiffening of the muscles immediately after death without being preceded by the stages of primary
relaxation.
·
The conditions necessary for its development are :
Somatic death must occur with extreme rapidity.
Person must be in state of great emotional tension.
Muscles must be in physical activity at that time.
Correct Answer. c
(51).
Legal age by which fetus is capable of independent existence:
a. 240 days
b. 230 days
c. 220 days
d. 210 days
Solution. Ans-51 : (d) 210 days
Ref: Read the text below.
Sol:
Criteria to diagnose live birth
1. Viable child (> 210 days according to law)
2. Well expanded lungs
3. Vital reaction in stump of umbilical cord
4. Presence of milk in stomach
Correct Answer. d
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(52).
Tattoos are useful in identifying :
a. Dead bodies
b. Living persons
c. Decomposed bodies
d. All of the above
Solution. Ans-52 : (d) All of the above
Ref: Read the text below.
Sol:
·
Tattoos are sometimes used byforensic pathologiststo help them identify burned, putrefied, or mutilated bodies.
·
Tattoo pigment is buried deep enough in the skin that even severe burns are not likely to destroy a tattoo
Correct Answer. d
(53).
Hair grows at the rate of :
a. 0.004 mm/day
b. 0.04 mm/day
c. 0.4 mm/day
d. 4.0 mm/day
Solution. Ans-53 : (c) 0.4 mm/day
Ref: Read the text below.
Sol:
·
Human hair typicallygrows at a rate of 0.4 mm/day, and this rate of growth is consistent across the head.
·
Several factors can influence both the rate and quality of hair growth, including diet, age, and general health
Correct Answer. c
(54).
Which of the following regarding exhumation is correct?
a. It is done in the presence of a police officer
b. It is done under the supervision of a medical officer
c. It can be done after a written order from a District Magistrate
d. All of the above
Solution. Ans-54 : (d) All of the above
Ref: Read the text below.
Sol:
Exhumation is the removal of remains from a grave.Exhumations are generally rare and can be traumatic for the bereaved family
involved. It is a requirement of a proposed exhumation that all close relatives of the deceased are contacted and sign to say they agree to
the proposed exhumation.
Exhumation of burial human remains and crematedremains will normally require a Home Office Licence issued from the Office of the
Secretary of State for the Home Office at the Department of Constitutional Affairs.
Reasons
Exhumations can occur for a number of reasons, including:
·
removal from the original grave site to a new grave acquired in the same or other cemetery
·
transfer from a public grave to a family grave
·
a Coroners instruction that requires further forensic examination of the deceased
·
It is done in the presence of a police officer
·
It is done under the supervision of a medical officer
·
It can be done after a written order from a District Magistrate
Correct Answer. d
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(55).
What is the time limit of exhumation in India?
a. 10 years
b. 30 years
c. 35 years
d. No time limit
Solution. Ans-55 : (d) No time limit
Ref: Read the text below.
Sol:
·
In India & England no time limit is fixed for disinterment of body
·
In France period is limited to 10 years & in Germany 30 years.
Correct Answer. d
(56).
A bullet find from a gun is not released. It is ejected out with the subsequent shot. It is known as:
a. Dumdum bullet
b. Rocketing bullet
c. Richochet bullet
d. Tandem bullet
Solution. Ans-56 : (d) Tandem bullet
Ref: Read the text below.
Sol:
Term
Feature
Tandem or Piggyback bullet
Two bullets are ejected one after another, when first bullet failed to leave the barrel & is
ejected by subsequently fired bullet.
Ricochet bullet
One which before striking the object aimed at, strikes some intervening object first, and
then after ricocheting & rebounding from this hits the object.
Yawning bullet
One which travel in irregular fashion instead of travelling nose on & cause a key hole
entry wound. Yaw means deviation between long axis of bullet and axis of path of bullet.
Correct Answer. d
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(57).
Brown-coloured urine is seen in the:
a. Nitric acid poisoning
b. Carbolic acid poisoning
c. Hydrochloric acid poisoning
d. Sulphuric acid poisoning
Solution. Ans-57 : (a) Nitric acid poisoning
Ref: Read the text below.
Sol:
Poison
Colour of lividity
Carbon monoxide (CO)
Bright cherry red
Burn, Cold, HCN
KCN, NaCN
Pink/Brown
Nitrites
Nitrates
Red Brown
Aniline
Potassium bicarbonate
Potassium Chlorate
Phosphorus (P)
Chocolate Brown
Dark Brown
H2S
Aniline, CO2
Opium
Blue-green
Deep blue
Black
Correct Answer. a
(58).
All are true about cadaveric spasm, except:
a. Appears instantaneously after death
b. Cannot be produced by any method after death
c. Molecular death does not occur
d. All the muscle of the body are involved
Solution. Ans-58 : (d) All the muscle of the body are involved
Ref: Read the text below.
Sol:
Cadaveric Spasm (Instantaneous rigor/cataleptic rigidity)
·
Rare condition caused by stiffening of the muscles immediately after death without being preceded by the stages of primary
relaxation.
·
The conditions necessary for its development are :
Somatic death must occur with extreme rapidity
Person must be in state of great emotional tension
Muscles must be in physical activity at that time
Correct Answer. d
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(59).
All the following methods are used in the detection of heavy metals, except:
a. Neutron activation analysis
b. Atomic absorption spectroscopy
c. Scanning electron microscope-energy dispersive X-ray analysis
d. Paraffin test
Solution. Ans-59 : (d) Paraffin test
Ref: Read thetext below.
Sol:
All the following methods are used in the detection of heavy metals, except Paraffin test.
Paraffin or dermal nitrate test is now an obsolete test, which was used earlier to determine if a suspect had discharged a firearm by
testing for presence of nitrate not heavy metals.
Correct Answer. d
(60).
All are true about LDL receptor except?
a. They are present on clathrin Co A ted pits on cell membrane
b. Taken by endocytosis.
c. They are present only at extra hepatic site
d. Increased cellular cholesterol downregulates the receptors
Solution. -NACorrect Answer. c
(61).
Amino acid residue having imino side chain is?
a. Lysine
b. Histidine
c. Tyrosine
d. Proline
Solution. Ans-61 : (d) Proline
Ref.:Read the text below
Sol :
The side chain of the proline makes pyrrolidine ring structure with the imino group.
Correct Answer. d
(62).
Glycine is present in?
a. Haemoglobin
b. Glutathione
c. Purine
d. All
Solution. Ans-62 : (d) All
Ref.:Read the text below
Sol :
Glycine is requiredfor the synthesis of haemoglobin, purine and glutathione.
Correct Answer. d
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(63).
A protein with M.W. of 100 KD is subjected to SDS-PAGE electrophoresis. The SDS PAGE electrophoresis pattern shows two widely
separated bands of 20 KD and 30 KD after addition of mercaptoethanol. The true statement regarding this will be?
a. Protein has undergone complete hydrolysis
b. Protein is a monomer of 20 KD and 30 KD protein
c. Protein is a dimer of 20 KD and 30 KD protein
d. Protein is a tetramer of 20 KD and 30 KD protein
Solution. Ans-63 : (d) Protein is a tetramer of 20 KD and 30 KD protein
Ref.:Read the text below
Sol :
Separation of the protein on the electrophoresis is based on charge and mass ratio.
Two band pattern of 20 and 30 KD denotes that two types of the protein having M.W of 20 KD and 30 KD is present in the sample.
As the total M.W. of the protein is 100 KD, it suggest that it is made up of 2×20KD and 2×30 KD.
Correct Answer. d
(64).
Chymotrypsinogen is a?
a. Zymogen
b. Carboxypeptidase
c. Transaminase
d. Clot lysing protein
Solution. Ans-64 : (a) Zymogen
Ref.:Read the text below
Sol :
The proteases are secreted as inactive zymogens; the active site of the enzyme is masked by a small region of the peptide chain that is
removed by hydrolysis of a specific Peptide bond.
Pepsinogen is activatedto pepsin by gastric acid and by activated pepsin (autocatalysis).
In the small intestine, trypsinogen, the precursor of trypsin, is activated by enteropeptidase, which is secreted by the duodenal epithelial
cells; trypsin can then activate chymotrypsinogen to chymotrypsin, proelastase to elastase, procarboxy-peptidase to carboxypeptidase,
and proaminopeptidase to aminopeptidase.
Correct Answer. a
(65).
The 40mm gap in between adjacent tropocollagen molecule in collagen which serve as the site of bone formation is occupied by which of
the following moiety?
a. Carbohydrate
b. Ligand moiety
c. Ca++
d. Fe+++
Solution. Ans-65 : (c) Ca++ as mineralization starts first in the gap between successive molecule of collagen.
Ref.:Read the text below
Sol :
The mechanisms involved in mineralization are not fully understood, but several factors have been implicated.
Alkaline phosphatase contributes to mineralization, but in itself is not sufficient.
Small vesicles (matrix vesicles) containing calcium and phosphate have been described at sites of mineralization, but their role is not
clear.
Type I collagenappears to be necessary, with mineralization being first evident in the gaps between successive molecules.
Recent interest has focused on acidic phosphoproteins such as bone sialoprotein acting as sites of nucleation.
These proteins contain motifs that bind calcium and may provide an initial scaffold for mineralization
Correct Answer. c
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24/71
(66).
Which of the following is a non-poisonous snake?
a. Viper
b. Krait
c. Sea snake
d. Rat snake
Solution. Ans-66: (d) Rat snake
Ref: Read the text below.
Sol:
·
Rat snakesare medium to large constrictorsthat can be found through a great portion of the northern hemisphere.
·
They feed primarily on rodentsand birdsand, with some species exceeding 3 m (10 feet), they can occupy top levels of some food
chains.
·
As with nearly all colubrids, rat snakes pose no threat to humans. Rat snakes are considered to be nonvenomous
Correct Answer. d
(67).
Forensic thanatology deals with :
a. Study of maggots swarming the body after death
b. Study of cooling of the body after death
c. Medico legal study of death
d. Suspended animation.
Solution. Ans-67 : (c) Medico legal study of death
Ref: Read the text below.
Sol:
·
Thanatologyis the academic, and often scientific, study of deathamong human beings.
·
It investigates the circumstances surrounding a person's death, the griefexperienced by the deceased's loved ones, and wider social
attitudes toward death such as ritual and memorialization.
Correct Answer. c
(68).
Kennedy phenomenon deals with:
a. Evaluation of exit and entry wounds
b. Evaluation of burn wounds
c. Effects of poisoning on stomach mucosa
d. Estimation of fetal age
Solution. Ans-68 : (a) Evaluation of exit and entry wounds
Ref: Read the text below.
Sol:
Kennedy Phenomenon : Surgical alteration or suturing of gunshot wounds create probems. The evaluation of the wound, whether it was
an entrance or an exit wound.
Correct Answer. a
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(69).
The most preferred side for postmortem collection of blood in alcohol poisoning is:
a. Femoral vein
b. Radial artery
c. Subclavian vein
d. Cardiac
Solution. Ans-69 : (a) Femoral vein
Ref: Read the text below.
Sol:
Collection of blood in Autopsy:
Before autopsy, 10 to 20 ml of blood can be drawn from the femoral vein by a syringe.
The flow may be increased by massaging the leg to drive blood proximally.
Correct Answer. a
(70).
What is the surest sign of death ?
a. Cadaveric lividity
b. Cadaveric spasm
c. Adipocere
d. Putrefaction
Solution. Ans-70 : (d) Putrefaction
Ref: Read the text below.
Sol: Putrefactionis one stage in the decompositionof the dead body; it can be described as the decomposition of proteins, which results in
the eventual breakdown of cohesion between tissues and liquification of most organs.
Correct Answer. d
(71).
Opium is the dried juice of which of the following plants ?
a. Capsicum annum
b. Datura fastuosa
c. Papaver somniferum
d. Cannabis sativa
Solution. Ans-71 : (c) Papaver somniferum
Ref.:Read the text below
Sol :
·
Opium(poppy tears, lachryma papaveris) is the dried latexobtained from the opium poppy(Papaver somniferum).
·
Opium contains up to 12% morphine,an alkaloid, which is frequently processed chemically to produce heroinfor the illegal drug
trade. The latex also includes codeineand non-narcotic alkaloidssuch as papaverine, thebaineand noscapine.
·
The traditional method of obtaining the latex is to scratch ("score") the immature seed pods (fruits) by hand; the latex leaks out and
dries to a sticky yellowish residue that is later scrapedoff.
The modern method is to harvest and process mature plants by machine. "Meconium" historically referred to related, weaker
preparations made from other parts of the poppy or different species of poppies
Correct Answer. c
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(72).
Hemodialyis can beuseful in the poisoning of :
a. Salicylic acid
b. Lithium,
c. ethylene glycol
d. All of above
Solution. Ans-72 : (d) All of above
Ref.: Read the text below
Sol :
The decision to initiate dialysis or hemofiltration in patients with renal failuredepends on several factors. These can be divided into acute
or chronic indications.
Indications for dialysis in the patient with acute kidney injuryare
1. Acidemiafrom metabolic acidosisin situations in which correction with sodium bicarbonate is impractical or may result in fluid
overload
2. Electrolyteabnormality, such as severe hyperkalemia, especially when combined with AKI
3. Intoxication, that is, acute poisoning with a dialyzable substance.: salicylic acid, lithium, isopropanol, Magnesium-containing
laxatives, and ethylene glycol
4. Overloadof fluid not expected to respond to treatment with diuretics
5. Uremiacomplications, such as pericarditis, encephalopathy, or gastrointestinal bleeding.
Correct Answer. d
(73).
Regarding HMP shunt, all of the following are true except ?
a. Occurs in the cytosol
b. No ATP is produced in the cycle
c. It is active in adipose tissues, liver, gonads
d. Oxidative phase generates NADPH and nonoxidative phase generates pyruvate
Solution. Ans-73 : (d) Oxidative phase generates NADPH and nonoxidative phase generates pyruvate
Ref:Read the text below
Sol :
- HMP shunt pathway occurs in the cytosol, in which glucose is oxidised. This is important for generation of ribose and NADPH.
It occurs in all the cell of the body, but in certain organs where reductive biosynthesis occurs, this pathway occurs to larger extent.
Such organs are liver, adipose tissue, adrenal glands, gonads, lactating mammary gland, etc.
It has got no energy value as it is not involved in ATP production.
Oxidative phase of this pathway is responsible for production of NADPH and nonoxidative phase is responsible for various intermediate
production, most important of which is ribose, involved in purine and pyrimidine biosynthesis.
Correct Answer. d
(74).
Decreased glycolytic activity impairs oxygen transport by haemoglobin Reason is :
a. Reduced energy production
b. Decreased production of 2, 3 bisphosphoglycerate
c. Reduced synthesis of haemoglobin
d. Low level of oxygen
Solution. Ans-74 : (b) Decreased production of 2, 3 bisphosphoglycerate
Ref:Read the text below
Sol :
- 2, 3 bisphosphoglycerate is produced during glycolysis within RBC in Rapport Leubering cycle.
- 2, 3 bisphosphoglycerate is an important allosteric modifier of haemoglobin 02 saturation and desaturation.
- This shifts O2 dissociation curve to the right.
- During decreased glycolytic activity decreased production of 2, 3 bisphosphoglycerate is responsible for increased O2 affinity to the Hb
and decreased dissociation from it leading to impaired O2 delivery at the tissue level.
Correct Answer. b
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(75).
In well fed state, acetyl Co A obtained from the diet is least used for the synthesis of ?
a. Palmityl CoA
b. Citrate
c. Acetoacetate
d. Oxalosuccinate
Solution. Ans-75 : (c) Acetoacetate
Ref:Read the text below
Sol :
Acetoacetate is the ketone bodywhich is synthesized during fatty acid oxidation in starvation.
In well fed state acetyl Co A is rather used for fatty acid synthesis. (palmityl Co A is a fatty acid).
Acetyl Co A is used in the TCA cycle for energy production, citrate and oxaloacetate are intermediate during TCA cycle.
Correct Answer. c
(76).
The activity of pyruvate carboxylase is dependent upon which positive allosteric effector ?
a. Succinate
b. Acetyl CoA
c. AMP
d. Isocitrate
Solution. Ans-76 : (b) Acetyl CoA
Ref:Read the text below
Sol :
Acetyl Co A derived fromfatty acid oxidation is an important positive allosteric activator of pyruvate carboxylase enzyme.
So in condition of starvation when there is excessive fatty acid oxidation, the acetyl Co A produced by fatty acid oxidation acts as a
stimulator of pyruvate carboxylase enzyme, stimulating gluconeogenesis.
Correct Answer. b
(77).
The biosynthesis of the enzyme pyruvate carboxylase is repressed by ?
a. Insulin
b. Cortisol
c. Glucagon
d. Epinephrine
Solution. Ans-77 : (a) Insulin
Ref:Read the text below
Sol :
Insulin diminishes gluconeogenesis.
It does so by repressing the activity of pyruvate carboxylase enzyme.
Correct Answer. a
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(78).
A musket is a :
a. Smooth bore weapon
b. Rifled weapon
c. Special type of shotgun in which the bore has been rifled
d. Smooth bore weapon which can be converted to a rifled one at the touch of a button
Solution. Ane-78 : (a) Smooth bore weapon
Ref.:Read the text below
Sol :
Musket is a smoothbored long-barelled old type Militarygun, with bayonet at the Muzle end.
It fires a single shot at high velocity.
Correct Answer. a
(79).
The effective range of shotgun is :
a. 10 – 15 yards
b. 20 – 30 yards
c. 30 – 40 yards
d. 60 – 80 yards
Solution. Ans-79 : (c) 30 – 40 yards
Ref.: Read the text below
Sol :
Effective range (sometimes also referred to as “killing range”) is the distanceupto which a victim can be hurt or killed.
It depends on a number of factors including degree of choking too. 30 – 40 yards is for a full choke gun.
Effective range is often contrasted with total or maximum range, which refers to the total distance upto the missile will travel (without
necessarily hurting the target).
Correct Answer. c
(80).
Helixometer is used for :
a. Finding the effective range of a gun
b. Finding the gauge or bore of a gun
c. Examining the interior of the barrel
d. Calculating the amount of gunpowder placed in a cartridge, so spurious cartridges could be separated
Solution. Ans-80 : (c) Examining the interior of the barrel
Ref.:Read the text below
Sol :
Helixometer was designed by John H. Fisher in 1925 for Bureau of Forensic Ballistics to examine the twist or helix of the rifling in a gun
barrel. It found limited application and fell into disuse.
Correct Answer. c
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(81).
Which of the following has highest electrophoretic mobility and least lipid content ?
a. Chylomicron
b. HDL
c. LDL
d. VLDL
Solution. Ans-81 : (b) HDL
Ref:Read the text below
Sol :
HDL also called alpha lipoprotein has got highest density and minimum lipid content. On electrophoresis HDL moves farthest on
electrophoretic plate towards cathode.
HDL = Alpha lipoprotein
VLDL = Prebeta lipoprotein
LDL = Beta lipoprotein
Correct Answer. b
(82).
Following statements are true regarding lipoproteins except ?
a. VLDL transport endogenous lipids
b. LDL transports lipid to the tissues
c. Increased blood cholesterol is associated with increased LDL receptors
d. Increased HDL is associated with decreased risk of coronary disease
Solution. Ans-82 : (c) Increased blood cholesterol is associated with increased LDL receptors
Ref:Read the text below
Sol :
- Increased cholesterol in the cholesterolpool of the cell is associated with decreases synthesis of cholesterol receptor and hence
decreases cholesterol uptake.
- Increase blood cholesterol isdirectly associated with increased concentration of cholesterol in the cholesterol pool of the cells which in
turn has got inverse effect on synthesis of cholesterol receptor on to the cells.
- Hence increase blood cholesterol is associated decreased cholesterol receptors on to the cell surface.
Correct Answer. c
(83).
NADPH is used in ?
a. Fatty acid synthesis
b. Ketone synthesis
c. Gluconeogensis
d. Glycolysis
Solution.
Ans-83 : (a) Fatty acid synthesis
Ref:Read the text below
Sol :
- NADPH is the source of reducing equivalentfor fatty acid synthesis.
- This NADPH is derived from 3 sources : A. HMP shunt pathway B. Isocitrate dehydrogenase, C. Malic enzyme.
- Ketone body synthesis, gluconeogenesis utilize NADH and glycolysis yields NADH.
Correct Answer. a
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(84).
The biosynthesis of the enzyme pyruvate carboxylase is repressed by ?
a. Insulin
b. Cortisol
c. Glucagon
d. Epinephrine
Solution. Ans-84 : (a) Insulin
Ref:Read the text below
Sol :
- Pyruvate carboxylaseis gluconeogenic hormone.
Insulin inhibits gluconeogenesis and has got inhibitory effect on pyruvate carboxylase.
Correct Answer. a
(85).
Protein purification and separation can be done by all except ?
a. Chromatography
b. Centrifugation
c. Electrophoresis
d. Densitometery
Solution. Ans 85: (d) Densitometery
Ref:Read the text below
Sol :
- Densitometery is the procedure which is used to measure the quantity of the protein ?
- Remember : For purification and separation following methods are used.
1. Chromatography
2. Dialysis
3. Centrifugation
4. Electrophoresis
5. Precipitation
Correct Answer. d
(86).
VMA is excreted excessively in the urine. What may be the diagnosis ?
a. Carcinoid syndrome
b. Phaeochromocytoma
c. Cushing’s syndrome
d. None
Solution. Ans-86: (b) Phaeochromocytoma
Ref:Read the text below
Sol :
- VMA is end product of catabolism of catecholamines.
- In phaeochromocytoma and neuroblastoma there is excessive synthesis of catecholamines which causes enhanced synthesis of VMA
and it’s excretion in the urine.
Correct Answer. b
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(87).
FIGLU is a metabolite of ?
a. Folic acid
b. Tyrosine
c. Histidine
d. Alanine
Solution. Ans-87 : (c) Histidine
Ref:Read the text below
Sol :
- FIGLU is a intermediate metabolite of histidine catabolism.
It is excreted in urine in increased quantity when there is folic acid deficiency as folic acid is required for further metabolism of FIGLU.
Correct Answer. c
(88).
False about tryptophan ?
a. Nonessential amino acid
b. Involved in serotonin synthesis
c. Involved in niacin synthesis
d. Involved in melatonin synthesis
Solution. Ans-88: (a) Nonessential amino acid
Ref:Read the text below
Sol :
- Tryptophan is an essential amino acid, as it’s synthesis in body does not take place.
- Tryptophan is involved in synthesis of melationin, serotonin, and niacin by various pathways.
Correct Answer. a
(89).
Kynurenine is formed from ?
a. Phenylalanine
b. Tryptophan
c. Tyrosine
d. Histidine
Solution. Ans-89: (bTryptophan
Ref:Read the text below
Sol :
- Kynurenin is synthesized fromtryptophan as a intermediate of it’s metabolic pathway.
- This pathway is also known as kynurenine pathway.
Correct Answer. b
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(90).
In arginase enzyme defect of urea cycle the amino acid excreted in the urine is ?
a. Lysine
b. Cystein
c. Bothe A and B
d. Arginine
Solution. Ans-90 : (c) Both A and B
Ref:Read the text below
Sol :
- In arginase defect in urea cycle arginine is accumulated which leads to excessive accumulation of arginine in the plasma and it’s
excretion in the urine initially in the PCT of the kidney but in the DCT this arginine is reabsorbed in exchange for lysine and cystine
which then gets excreted in the urine.
- So in arginase defect amino acid accumulated is arginine but the one that is excreted is lysine and cystein.
Correct Answer. c
(91).
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. Ans-91 : (a) Lung
Ref:Read the text below
Sol: Blast caused by an explosionis 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 organshaving an interface with air are (ii) middle ear and (iii)gastrointestinal tract. These two suffer greatly in
explosions.
Correct Answer. a
(92).
Which of the following tissues is most resistant to electric current?
a. Muscle
b. Skin
c. Bone
d. Blood
Solution. Ans-92 : (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
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(93).
“Retraction balls“ are seen after injuries to:
a. Liver
b. Spleen
c. Lungs
d. Brain
Solution. Ans-93 : (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
(94).
Pond fracture is:
a. A simple dent of the skull
b. Caused by falling into a deep pond
c. Mostly seen in bones of infants
d. Both (a) and (c) above
Solution. Ans-94 : (d) Both (a) and (c) above
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. d
(95).
Which of the following statements regarding “Bumper Fractures” is true?
a. Usually the tibia is fractured
b. Usually the fibula is fractured
c. Usually calcaneum is fractured
d. Invariably all three, i.e. tibia, fibula and the calcaneum are fractured
Solution. Ans-95 : (a) Usually the tibia is fractured
Ref:Read the text below
Sol:
·
Usually tibia is fractured, although very rarely fibula may also fracture. The fractured fragment of tibia is triangular in shape.
·
The base of the triangular fragment indicates the site of impact, and the apex points in the direction in which the vehicle was
travelling.
·
Occassionally there are no fractures, or even any evidence of external injury, but there may be internal haemorrhage in the calves.
Correct Answer. a
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(96).
In which part of India, coroner’s system of inquest is followed?
a. In Madras
b. Nowhere in India
c. In Calcutta
d. In Mumbai
Solution. Ans-96 : (b) Nowhere in India
Ref:Read the text below
Sol:
·
Coroner’s inquestis nowhere held in India now. The last city to abolish it was Mumbai, where it was abolished on 29th July 1999.
Currently coroners’ inquests are held in England.
·
They evolved in England over eight centuries having been formally established in 1194.
·
The role has changed from that of a form of medieval tax gatherer into becoming an independent judicial officer charged with the
investigation of sudden, violent or unnaturaldeath.
Coroner is called so because he was traditionally appointed directly by the king (Crown).
Correct Answer. b
(97).
There are five basic categories of drugs that are controlled by the Controlled Drugs and Substances Act, second schedule includes
a. Heroin
b. Morphine
c. Cocaine
d. Marijuana
Solution. Ans-97 : (d) Marijuana
Ref:Read the text below
Sol:
Controlled Drugs And Substances Act
·
There are five basic categories of drugs that are controlled by the Controlled Drugs and Substances Act.These categories are all
listed in schedules attached to the Act. Drugs listed in schedules 4 and 5, which include substances such as steroids and barbituates, can
be possessed but must be obtained with a prescription, and are illegal to import, export or traffic, or to possess for the purpose of
trafficking.
·
Schedules 1, 2, and 3 contain drugswhich are illegal to possess. There are literally hundreds of drugs in these schedules.
·
The first schedule includes drugs such as heroin, morphine, cocaine, codeine and many other similar substances.
·
The second schedule contains cannabis (marijuana) products.
·
The third schedule contains drugs like amphetamines, LSD, psilocybin and methamphetamines
Correct Answer. d
(98).
The term “ Panchanama” refers to
a. Order by a police officer compelling the witness to appear in police station
b. A local body of five people deciding criminal cases
c. Inquest report
d. Order by the court compelling the witness to appear before it
Solution. Ans-98 : (c) Inquest report
Ref:Read the text below
Sol:
·
Traditionally a police officer investigating a murder case, used to record information from five (Hindi: Panch) witnesses.
·
These witnesses were preferably some respectable people from the locality who were supposed to speak the truth.
·
The report thus prepared was called panchanama.
·
Now usually statements from two witnesses are recorded.
Correct Answer. c
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(99).
Asterixis is seen in
a. Arsenic trioxide
b. Arsenic phosphide
c. Red phosphorus
d. Yellow phosphorus
Solution. Ans-99 : (d) Yellow phosphorus
Ref:Read the text below
Sol:
·
Asterixis is the namegiven to flapping tremor of the hands. It is seen in third stage of acute phosphorus poisoning.
Correct Answer. d
(100).
Where does a lacerated wound appear like an incised wound?
a. Abdomen
b. Thorax
c. Hand
d. Forehead
Solution. Ans-100 : (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
(101).
When the semen of donor is mixed with that of the infertile husband, the procedure is known as
a. AIH
b. AID
c. AIHD
d. None of the above
Solution. Ans-101 : (c) AIHD
Ref:Read the text below
Sol: AIHD stands for Artificial Inseminationby Husband and Donor. It is sometimes also known as Artificial Insemination Mixed (AIM).
This method offers no special advantage over AID, except that the husband may believe that he could be the father of the child.
·
In fact AIHD has been criticized since semen from the infertile husband may contain antibodies which could interfere with normal
sperm function.
However studies have not confirmed this. In some foreign jurisdictionsAIHD is outlawed. Indian law is silent on this.
Correct Answer. c
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(102).
An 18 year old unmarried female has been brought to a doctor. On examination the hymen is found to be intact, labia majora are
separated and flabby, the labia minora are cutaneous in appearance and separated, the fourchette is torn and the vagina is roomy and
enlarged. Most probably the woman is:
a. A true virgin
b. A false virgin
c. Sterile
d. In premenstrual stage
Solution. Ans-102 : (b) A false virgin
Ref:Read the text below
Sol:
·
Traditionally it has been believed that if the hymen of a female is intact, she is a virgin, i.e. she has not experienced sexual
intercourse.
·
However this has proved to be false on a number of occassions.
·
There are several cases on record where a female has had sexual intercourse and yet her hymen was intact. Hymen has been found
intact in women coming for delivery, and even in prostitutes.
·
The following definitions are important:
True virgin: A woman who has never experienced sexual intercourse. Her hymen may or may not be intact.
False virgin: A woman who has experienced sexual intercourse, but her hymen is intact.
Correct Answer. b
(103).
Statutory rape is:
a. Rape on a girl below 16 years
b. Rape of another person’s wife
c. Intercourse with one’s own wife without her consent
d. Consensual Intercourse with a girl
Solution. Ans-103 : (d) Consensual Intercourse with a girl
Ref:Read the text below
Sol:
·
In some common law jurisdictions, statutory rape is sexual activity in which one person is below the age required to legally consent to
the behavior.
·
Although it usually refers to adults engaging in sex with minors under the age of consent,it is a generic term, and very few
jurisdictions use the actual termstatutory rape in the language of statutes.
Different jurisdictions use many different statutory terms for the crime, such assexual assault (SA), rape of a child (ROAC), corruption of
a minor (COAM),unlawful sex with a minor (USWAM),carnal knowledge of a minor (CKOAM),unlawful carnal knowledge (UCK), sexual
battery[4] or simply carnal knowledge
Correct Answer. d
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(104).
If a person is caught with the amount of marijuana more than 30 grams, the maximum penalty is
a. 1 years
b. 5 years
c. 7 years
d. 10 years
Solution. Ans-104 : (b) 5 years
Ref: Read the text below
Sol:
In addition, the general circumstances surrounding the offence and the personal circumstances of the person being sentenced are
considered by a sentencing judge. If a person has a prior criminal record, especially a drug record, the sentence will probably be more
harsh than if the person has no previous involvement with the law. Consider the maximum sentences which can be imposed for
possession of Marijuana as an example:
ØMarijuana (schedule 2)- up to $1,000.00 and six months in jail if the amount is less than 30 grams (or one (1) gram of hash);
ØIf the amount of marijuana is more than 30 grams (or more than one (1) gram of hash) the maximum penalty is five (5) years less a day
in prison, if the Crown proceeds by indictment. Maximums are less if the Crown proceeds by summary conviction.
ØIn the case of schedule one (1) drugs (such as heroin, cocaine or morphine) - the maximum penalty if the Crown proceeds by indictment
is seven (7) years in jail. If the Crown proceeds by summary conviction, the maximum penalty is less.
ØFor schedule three (3) drugs - the maximum penalty by indictment is three (3) years and less for summary conviction.
Correct Answer. b
(105).
All of the following are cognizable and non bailable offences, except
a. Murder
b. Rape
c. Attempted suicide
d. Up rise against nation
Solution. Ans-105 : (c) Attempted suicide
Ref:Read the text below
Sol:
·
Murder, rape and uprise against nation are both cognizable and non bailable offences. But attempted suicide is bailable.
Correct Answer. c
(106).
In hostile witness scenario leading questions are asked in
a. Examination in chief
b. Cross examination
c. Dying declaration
d. Re direct examination
Solution. Ans-106: (a) Examination in chief
Ref:Read the text below
Sol:
·
A hostile witness is the one who changes or conceals a part of the statemnet he has given earlier to the police at the time when he is
present in the court to give evidence.
·
In this scenario his own lawyer gains the power to cross question him and thus can ask leading questions to him.
Correct Answer. a
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(107).
All the following are the defences available for the doctor in cases of negligence, except
a. Time limitation
b. No fees accepted
c. Therapeutic misadventure
d. Res judicata
Solution. Ans-107 : (b) No fees accepted
Ref:Read the text below
Sol:
Following are the defences available for the doctor in cases of negligence
ØTime limitation
ØTherapeutic misadventure
ØRes judicata
Correct Answer. b
(108).
If a person is not heard for…….years by his relatives or friends, he is presumed to be dead
a. 3
b. 7
c. 10
d. 12
Solution. Ans-108 : (b) 7
Ref:Read the text below
Sol:
Currently, the law generally assumes a person is dead if, after seven years:
·
There has been no evidence that they still live.
·
The people most likely to have heard from them have had no contact.
·
Inquiries made of that person have had no success
Correct Answer. b
(109).
Which of the following constitutes Grievous injury
a. Incised wound over the scalp
b. Lacerated wound over the scalp
c. Fracture of radius
d. Hospitalization for 15 days
Solution. Ans-109 : (c) Fracture of radius
Ref:Read the text below
Sol:
Grievous hurt.-- The following kinds of hurt only are designated as" grievous":·
First.- Emasculation.
·
Secondly.- Permanent privation of the sight of either eye.
·
Thirdly.- Permanent privation of the hearing of either ear.
·
Fourthly.- Privation of any member or joint.
·
Fifthly.- Destruction or permanent impairing of the powers of any member or joint.
·
Sixthly.- Permanent disfiguration of the head or face.
·
Seventhly.- Fracture or dislocation of a bone or tooth.
·
Eighthly.- Any hurt which endangers life or which causes the sufferer to be during the space of twenty days in severe bodily pain, or
unable to follow his ordinary pursuits.
Correct Answer. c
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(110).
All of the following is true about sickle cell disease except?
a. Single nucleotide change results in change of glutamine to valine
b. RFLP results from a single base change
c. Sticky patch is generated as a result of replacement of nonpolar residue with a polar residue
d. HbS confers resistance against malaria in heterozygotes
Solution. Ans-110 : (c) Sticky patch is generated as a result of replacement of nonpolar residue with a polar residue
Ref.:Read the text below
Sol :
- In sickle cell disease valine (nonpolar) appears at the place of glutamic acid (polar) so the above statement is wrong.
- All other statements are right, RFLP pattern is observed due to polymorphism or mutation.
Correct Answer. c
(111).
The amino acid which serve as a carrier of ammonia from skeletal muscle to the liver is?
a. Alanine
b. Methionine
c. Arginine
d. Glutamine
Solution. Ans-111 : (a) Alanine Ref.: Read the text below Sol :
- During exercise excessive alanine is produced from the skeletal muscle, either due to protein breakdown or due to transamination of
pyruvate produced via glycolysis.
- This alanine is carried to the liver where it synthesizes glucose via gluconeogenesis.
Correct Answer. a
(112).
Phenylalanine is the precursor of all of the following except?
a. Tyrosine
b. Epinephrine
c. Thyroxine
d. Melatonin
Solution. Ans-112: (d) Melatonin Ref.: Read the text below Sol :
- Melatonin is synthesized from tryptophan.
- Tyrosine, epinephrine, and thyroxin are all synthesized from phenylalanine.
Correct Answer. d
(113).
Which of the following combination is wrong?
a. Phenylalanine – niacin
b. Tryptophan – serotonin
c. Phenylalanine – melanin
d. Tyrosine – epinephrine
Solution. Ans-113 : (a) Phenylalanine – niacin Ref.: Read the text below Sol :
- Niacin is synthesized from tryptophan and not from tyrosine. So this combination is wrong.
- Rest other combinations are correct because respective compounds are synthesized from corresponding amino acids.
Correct Answer. a
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(114).
All of the following are trypsin inhibitors, except?
a. Alpha – 1 – antitrypsin
b. Alpha – 1- anti protease
c. Enterokinase
d. Egg white
Solution. Ans-114 : (c) Enterokinase Ref.: Read the text below Sol : Inhibitors of trypsins are
- Alpha-1-antitrypsin (antiprotease)
- Egg white
- Raw soybean
- Human and bovine colostrums
- di-isopropyl fluro phosphate
- activators of trypsin enterokinase, calcium, trypsin as such are activators of trypsin
Correct Answer. c
(115).
Collagen triple helix structure is not found in the?
a. Cytoplasm
b. Golgi apparatus
c. Lumen of endoplasmic reticulum
d. Intracellular vesicle
Solution. Ans-115 : (a) Cytoplasm Ref.: Read the text below Sol :
- In the cytosol the newly formed collagen is pro chain (single strand), the triple helical structure is seen only in RER, Golgi apparatus
and vesicle which is formed thereafter.
- Fully organized collagen fibril is released in the extracellular space.
Correct Answer. a
(116).
In which one of the following tissue glucose transport in the cell is enhanced by insulin?
a. Brain
b. Lens
c. RBC
d. Adipose tissue
Solution. Ans-116: (d) Adipose tissue Ref.: Read the text below Sol :
- Major tissues in which glucose transport require insulin are muscle and adipose tissue.
- The metabolism in the liver responds to insulin, but hepatic glucose transport is determined by blood glucose concentration, and dose
not require insulin.
Correct Answer. d
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(117).
To produce insulin for therapeutic purpose in bacteria involves which of the following as initial material?
a. mRNA from beta pancreatic cell of human
b. Genomic DNA from beta pancreatic cell of human
c. mRNA from lymphocyte of human
d. Genomic DNA from lymphocyte of human.
Solution. Ans-117: (a) mRNA from beta pancreatic cell of human Ref.: Read the text below Sol :
- Restriction fragment of the DNA of an organism is collected in DNA library after recombinant DNA technology.
- Joining of two different DNA via DNA ligase enzyme is known as recombinant DNA technology.
- Depending upon the source of the DNA which is used for insertion in the vector DNA gene library is of two types :
Correct Answer. a
(118).
Methylation of bases in DNA usually?
a. Facilitates the binding of transcription factor to the DNA
b. Makes a difference in the activity only if it occurs in an enhancer region
c. Inactivates DNA for transcription
d. Results in increased production of product in whatever gene it is methylated.
Solution. Ans-118 : (c) Inactivates DNA for transcription Ref.: Read the text below Sol :
- Methylated DNA is the inactive form of the DNA.
- Option A and option D says that the methylation enhances the rate of transcription which is a wrong statement.
- Alteration in enhancer region may make a difference, but alteration in the promoter region would certainly make the difference.
Correct Answer. c
(119).
All are true about LDL receptor except?
a. They are present on clathrin coated pits on cell membrane.
b. Taken by endocytosis
c. They are present only at extra hepatic site.
d. Increased cellular cholesterol down regulates the receptors
Solution. Ans-119 : (c) They are present only at extra hepatic site. Ref.: Read the text below Sol :
- LDL receptor is present on liver as well as on extrahepatic tissue.
- LDL (apo B-100, E) receptors occur on the cell surface in pits that are coated on the cytosolic side of the cell membrane with a protein
called clathrin.
- The glycoprotein receptor spans the membrane, the B-100 binding region being at the exposed amino terminal end.
Correct Answer. c
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(120).
Encircle the material which is not involved in protein translation in eukaryotes?
a. RNA polymerase
b. Aminoacylt – RNA
c. Ribosomes
d. Peptidyl transferase
Solution. Ans-120 : (a) RNA polymerase Ref.: Read the text below Sol :
- RNA polymerase enzyme is involved in transcription process, not in translation process.
- Aminoacyl t RNA is needed during the process of translation to carry the amino acid at the surface of the ribosome for the attachment
in the elongating polypeptide chain.
- Ribosome is the organell on which the protein synthesis takes place.
Correct Answer. a
(121).
The 40 nm gap in between adjacent tropocollagen molecule in collagen which serve as the site of bone formation is occupied by which of
the following moiety?
a. Carbohydrate
b. Ligand moiety
c. Ca++
d. Fe+++
Solution. Ans-121 : (c) Ca++ Ref.: Read the text below Sol :
- “The mechanisms involved in mineralization are not fully understood, but several factors have been implicated.
- Alkaline phosphatase contributes to mineralization, but in itself is not sufficient.
- Small vesicles (matrix vesicles) containing calcium and phosphate have been described at sites of mineralization, but their role is not
clear.
- Type I collagen appears to be necessary, with mineralization being first evident in the gaps between successive molecules.”
Correct Answer. c
(122).
The predominant isoenzyme of LDH in the cardiac muscle is?
a. LDH – 1
b. LDH – 2
c. LDH – 3
d. LDH – 5
Solution. Ans-122 : (a) LDH – 1 Ref.: Read the text below Sol :
- In normal plasma LDH-2 is more in concentration than LDH – 1.
- In myocardial infarction level of LDH – 1 increases and this leads to altered ratio of LDH.
- It means LDH – 1 > LDH – 2.
- This altered ratio of the LDH is known as flipped pattern.
Correct Answer. a
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(123).
Aminoacyl – t- RNA is required for all except ?
a. Hydroxyproline
b. Methionine
c. Cystine
d. Lysine
Solution. Ans-123 : (a) Hydroxyproline
Ref.: Read the text below
Sol :
- Hydroxyproline is a derived amino acid and there is no t-RNA available to transport derived amino acids: hydroxyproline and
hydroxylysine are two examples of derived amino acid for which there is no t-RNA available.
Correct Answer. a
(124).
True about Cyanide is ?
a. Only minimally inhibits the ETC because cytochrome oxidase is the terminal component of the chain
b. Inhibits mitochondrial respiration but energy production is unaffected
c. Also binds the copper of cytochrome oxidase
d. Bind to Fe+++ of cytochrome a3
Solution. Ans-124 : (d) Bind to Fe+++ of cytochrome a3
Ref.: Read the text below
Sol :
- That is why methhemoglobin is an effective antidote since it also has Fe+++.
- Respiration and energy production is the coupled process, so inhibition of one inhibits the other as well.
- Cu is an important part of cytochrome oxidase, but cyanide does not bind it.
Correct Answer. d
(125).
Which of the following statement are true regarding RDA ?
a. RDA is statistically defined as two standard deviation above the estimated average requirement (EAR)
b. RDA is statistically defined as equal to the EAR
c. RDA is statistically defined as equal to the adequate intake
d. RDA is defined as the recommended minimum requirement
Solution. Ans-125 : (a) RDA is statistically defined as two standard deviation above the estimated average requirement (EAR)
Ref.: Read the text below
Sol :
- RDA (Recommended dietary allowance) :
- Amount of nutrient estimated to meet the nutrient requirement of 97% to 98% of healthy individual in an age and gender group.
- RDA = EAR + 2SDEAR
Correct Answer. a
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(126).
BMR, true is ?
a. Is not influenced by energy intake
b. Increases in response to starvation
c. May decrease up to 50% during period of starvation
d. Not responsive to change in the hormonal level
Solution. Ans-126 : (c) May decrease up to 50% during period of starvation
Ref.: Read the text below
Sol :
-This is a part of survival mechanism in starvation.
- Reduced energy intake reduces BMR and increased energy intake increases BMR.
- BMR changes in response to hormone level. Thyroid hormone, cortisol etc increases the BMR.
Correct Answer. c
(127).
All are true about nucleotide excision repair except ?
a. Removal of damaged bases occurs only on one strand of the DNA
b. It removes thymine dimmers generated by uv light
c. Requires ligase and polymerase
d. Only the damaged nucleotide are removed
Solution. Ans-127 : (d) Only the damaged nucleotide are removed
Ref.: Read the text below
Sol :
- Cuts are made several nucleotide on either side of the damaged base. The intact strand acts as a template strand for the repair. Many
lesions are removed by this nucleotide excision repair.
- Ultraviolet induced thymine dimmers are only one example of such lesions. To fill the gap, polymerase and ligase activity is required.
Correct Answer. d
(128).
True statement regarding base excision repair is ?
a. Used only for the bases which are deaminated
b. Uses enzyme called as DNA glycosylase which generate abasic suger site
c. Removes about 10 – 15 nucleotide
d. Recognizes a bulk lesion
Solution. Ans-128 : (b) Uses enzyme called as DNA glycosylase which generate abasic suger site
Ref.: Read the text below
Sol :
- This is the first step of base excision repair. Bases modified by deamination, methylation, or by other chemical modification are removed
by this kind of repair system.
Correct Answer. b
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(129).
If consecutive nucleotides code for an amino acid, how many amino acid can be theoretically coded by nucleic acid?
a. 4
b. 64
c. 16
d. 256
Solution. Ans-129 : (d) 256
Ref.: Read the text below
Sol :
- If there would have been two nucleotide in each codon the chance of their varied combination will be 42 i.e. 16, and three nucleotide in
each codon means there will occur 43 i.e. 64 codons are possible.
- According to the above calculation, if there occurs 4 nucleotide in a codon, the chance of there varied combination will be 44 i.e. 256.
- So the answer of above question will be (D) 256.
Correct Answer. d
(130).
Squalene is the intermediate product during synthesis of?
a. VLDL
b. Cholesterol
c. Tachysterol
d. Lanosterol
Solution. Ans-130 : (b) Cholesterol
Ref: Read the text below
Sol :
- Squalene is 30 carbon structure which is formed as an intermediate during synthesis of cholesterol.
- From squalene cholesterol is formed after removal of 3 CH3 group.
Correct Answer. b
(131).
In the body, metabolism of 10 grams of protein would produce approximately
a. 1 K calorie
b. 41 K calories
c. 410 K calories
d. 41 Calories
Solution. Ans-131 : (b) 41 K calories
Ref.: Park, - 547
Sol :
- The dietary sources of energy are proteins, fats and carbohydrates. They supply energy at the following rates :
- Proteins - 4 kCal/g (or 17kj)
- Fats – 9 kCal/g (or 37 kj)
- Carbohydrates – 4 kCal/g (or 17Kj)
- Hence, energy yield from 10 grams of protein will be 40 kCals.
Correct Answer. b
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(132).
Refsum’s disease is a rare neurological disorder caused by
a. Accumulation of glutamic acid
b. Accumulation of glycine
c. Accumulation or phytanic acid
d. Accumulation of phenylalanine
Solution. Ans-132 : (c) Accumulation or phytanic acid
Ref.: Harrison’s, - 2664.
Sol :
- Refsum’s disease is due to accumulation of phytanic acid due to deficiency of enzyme phytanic acid hydroxylase.
- It is an autosomal recessive condition characterized by mixed motor and sensory neuropathy, palpable nerves, ataxia, retinitis
pigmentosa, night blindness, sensorineural deafness, elevated CSF protein, and age of onset before 20 years.
- Histopathologically severe onion bulb formation is seen.
Correct Answer. c
(133).
Which one of the following pairs of lipids and related compounds exhibit opposite biological activities ?
a. 5-HPETE and leukotriene D4
b. Cholic acid and lithocholic acid
c. Thromboxane A2 and prostocyclin (PG12
d. Acetone and beta-hydroxybutyrate
Solution. Ans-133 : (c) Thromboxane A2 and prostocyclin (PG12)
Ref.: Lippincott’s Biochemistry 2nd edn. Pg. 186.
Sol :
Thromboxane A2
Prostacyclin
Produced primarily by platelets
Produced primarily by endothelium of vessels
Promotes platelet aggregation
Inhibits platelet aggregation
Decreases formation of cAMP
Increases formation of cAMP
Vasoconstriction
Vasodilatation
Mobilizes intracellular calcium
Contraction of smooth muscle
Correct Answer. c
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(134).
The cost in high-energy phosphate bonds for the formation of 1 mole of glucose from pyruvate is
a. 8 moles of ATP
b. 6 moles of ATP
c. 4 moles of ATP
d. 2 moles of ATP
Solution. Ans-134 : (b) 6 moles of ATP
Ref.: Vasudevan Biochemistry - 95.
Sol :
- Energy requirement for gluconeogenesis
Correct Answer. b
(135).
Which of the statements best characterize glucose ?
a. It usually exits in furanose form
b. It is a ketose
c. It forms part of the disaccharide sucrose
d. It is oxidized to form sorbitol
Solution. Ans-135 : (c) It forms part of the disaccharide sucrose
Ref.: Lippincott’s Biochemistry, 2nd edn. Pg. 120, 130.
Sol :
- Glucose is reduced by aldose reductase to form sorbitol
- Glucose is a monosaccharide (hexose). It has an aldehyde group (CHO) and hence is an aldose (aldohexose).
- Fructose is a ketohexose.
- Glucose exists in biologic systems as a pyranose ring.
- Sucrose consists of glucose + fructose.
Correct Answer. c
(136).
Which of the following enzymes catalyze reactions in the biosynthesis of both catecholamines and indoleamines (serotonin) ?
a. Dopamine beta-hydroxylase
b. Phenyl ethanolamine N-methyl transfer
c. Aromatic amino acid decarboxylase
d. Tryptophan hydroxylase
Solution. Ans-136 : (c) Aromatic amino acid decarboxylase Ref.: Lippincott’s Biochemistry - 266-67 Sol :
- In synthesis of catecholamines, dopa is decarboxylated to dopamine (dopa-decarboxylase). Catecholamines are derived from tyrosine.
- In synthesis of serotonin, 5-hydroxytryptophan is decarboxylated to serotonin (decarboxylase). Serotonin is derived from tryptophan.
Correct Answer. c
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(137).
Immediate precursor in the production of acetoacetate is?
a. Acetoacetyl Co A
b. Hydroxyl methyl glutaryl Co A
c. Malonyl Co A
d. Iso valeric acid
Solution. Ans-137 : (b) Hydroxyl methyl glutaryl Co A
Ref: Read the text below
Sol :
- HMG Co A is the immediate precursor of acetoacetate which is the 1st ketone body formed.
- β-(OH) butyrate is formed once β–(OH) butyrate dehydrogenase acts on acetoacetate.
- β-(OH) butyrate is most abundant (predominant) ketone body formed.
Correct Answer. b
(138).
LDL receptor in liver can be detected by?
a. Apo B 100 and apo E
b. Apo B 100 and apo A
c. Apo E
d. Apo E and apo A
Solution. Ans-138 : (a) Apo B 100 and apo E
Ref: Read the text below
Sol :
- LDL receptors are present on the liver and are involved in removal of LDL particle formed in the circulation.
- They recognize Apo B-100 and Apo E.
- So LDL receptors are not only involved in removal of LDL particles having Apo B-100, they also can remove chylomicron remnant
having ApoE.
Correct Answer. a
(139).
False about insulin action?
a. Glycogen synthesis
b. Glycolysis
c. Lipogenesis
d. Ketogenesis
Solution. Ans-139 : (d) Ketogenesis
Ref: Read the text below
Sol :
- Insulin inhibits lipolysis by inhibiting hormone sensitive lipase. Thus insulin prevents ketogenesis.
- Insulin stimulates glycolysis by activating PFK – 1
- Insulin activates glycogenesis by activating glycogen synthase.
- Also it stimulates fatty acid synthesis by activating acetyl Co A caroboxylase.0
Correct Answer. d
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(140).
In synthesis of fatty acid energy is supplied by?
a. NAD
b. FAD
c. GTP
d. NADPH
Solution. Ans-140 : (d) NADPH
Ref: Read the text below
Sol :
- NADPH is involved in various reductive biosynthesis.
- NADPH is generated in HMP shunt pathway.
- In fatty acid synthesis each cycle of fatty acid synthesis require 2 molecule o NADPH
Correct Answer. d
(141).
Which one of these has the least density?
a. Chylomicron
b. HDL
c. LDL
d. VLDL
Solution. Ans-141 : (a) Chylomicron
Ref: Read the text below
Sol :
- Density of various lipoproteins are determined by lipid and protein content of the particle
- More the lipid content more lighter the particle is, chylomicron is lightest of all the lipoprotein.
Correct Answer. a
(142).
The biosynthesis of the enzyme pyruvate carboxylase is repressed by?
a. Insulin
b. Cortisol
c. Glucagon
d. Epinephrine
Solution. Ans-142 : (a) Insulin
Ref: Read the text below
Sol :
- Pyruvate carboxylase is gluconeogenic hormone.
- Insulin inhibits gluconeogenesis and has got inhibitory effect on pyruvate carboxylase.
Correct Answer. a
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(143).
Rate limiting enzyme of cholesterol biosynthesis is?
a. HMG Co A reductase
b. HMG Co A synthase
c. 7 alpha hydroxylase
d. Phosphofuctokinase
Solution. Ans-143 : (a) HMG Co A reductase
Ref: Read the text below
Sol :
- HMG Co A reductase is the rate limiting enzyme for cholesterol synthesis.
- HMG Co A reductase enzyme is both under covalent modification as well as under allosteric regulation.
- Insulin and thyroid hormone increases the activity of HMG Co A reductase and mevalonate, cholesterol, glucagon, glucocorticoids
inhibit the activity of HMG Co A reductase enzyme.
Correct Answer. a
(144).
Hypolipidemic agents acts on?
a. HMG Co A synthetase
b. HMG Co A reductase
c. HMG Co A mutase
d. HMG Co A lyase
Solution. Ans-144 : (b) HMG Co A reductase
Ref: Read the text below
Sol :
- Hypolipidemic drugs (statin) when given to reduce cholesterol content in the circulation, do so via inhibiting endogenous synthesis of
cholesterol by acting on HMG Co A reductase enzyme.
Correct Answer. b
(145).
Bile acids are derived from ?
a. Cholesterol
b. Fatty acid
c. Bilirubin
d. Proteins
Solution. Ans-145 : (a) Cholesterol
Ref: Read the text below
Sol :
- Primary bile acids (cholic acid and chenodeoxycholic acids) are synthesized from cholesterol by action of 7 α- hydroxylase and 12 α
Hydroxylase in peroxisosome of liver.
Correct Answer. a
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(146).
Which biochemical reaction does not occur in mitochondria?
a. Kreb’s cycle
b. Urea cycle
c. Gluconeogenesis
d. Fatty acid synthesis
Solution. Ans-146 : (d) Fatty acid synthesis
Ref: Read the text below
Sol :
- Fatty acid synthesis occurs in the cytoplasm.
- In cytoplasm there occurs fatty acid synthase complex, which is a dimer and where occurs fatty acid synthesis.
- Kreb’s cycle occurs entirely within the mitochondria.
- Gluconeogenesis and urea cycle occurs partly in mitochondria and partly in cytoplasm.
Correct Answer. d
(147).
One week infant, having classic phenylketonuria, which of the following statement is correct?
a. Tyrosine is nonessential amino acid for this infant
b. High levels of phenylpyruvate is appearing in the urine
c. Therapy must begin within first year of life
d. Diet therapy should be discontinued once infant reaches adulthood.
Solution. Ans-147 : (b) High levels of phenylpyruvate is appearing in the urine
Ref: Read the text below
Sol :
- Tyrosine becomes essential amino acid for this infant as phenylalnine can not be converted to tyrosine.
- Phenylalanine is converted to phenylpyruvate, phenylacetate and phenyllactate in this disorder.
- Therapy of phenylalanine restricted diet should be started within 7 to 10 days of life to avoid mental retardation.
- Adult PKU patient shows deterioration of attention and thinking process after discontinuation of diet, so life long restriction of dietary
phenylalanine is recommended.
Correct Answer. b
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(148).
Prokaryotic promoter of transcription situated 35 bp upstream is:
a. TATA box
b. Goldberg hogness box
c. CAAT box
d. TGG box
Solution. Ans-148 : (d) TGG box
Ref.: Read the text below
Sol :
- Bacterial promoter situated 35 bp upstream to the starting site of transcription is TGG box
- Promoters are short conserved sequences in the coding strand of DNA that specifies the start of transcription
Bacterial promoters :
- TATA box of pribnow box
Situated 10 bp upstream
- TGG box
Situated 35 bp upstream
- Eukaryotic promoters :
- Goldberg hogness box
Situated 25 bp upstream
- CAAT box
Situated 25 bp upstream
- CAAT box
Situated 70 bp upstream GC rich region
- The DNA strand that is transcribed into a RNA molecule is : Template/sense strand/n on coding/’+’ strand
- Opposite strand is : Coding/antisense/non template/ ‘-‘ strand Sigma subunit of RNA polymerase recognizes promoter site
Correct Answer. d
(149).
All are true about fibrous proteins except:
a. They have mainly structural functions
b. They have an axial ratio of < 3
c. Elastin is a fibrous protein
d. They have negligible water solubility
Solution. Ans-149 : (b) They have an axial ratio of < 3
Ref.: Read the text below
Sol :
- Axial ratio is the ratio of length to breadth.
- Fibrous proteins possess axial ratio of 10 or more.
Fibrous Proteins :
- Structural proteins
- Minimum water solubility
- Axial ratio > 10
- Eg. Collagen, Keratin, elastin
Globular proteins
- Spherical shape
- Axial ratio < 3
- Dynamic functions
- Eg. Albumin, globulin, most enzymes
Correct Answer. b
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(150).
Which of the following does not contain β–alanine?
a. Pantothenic acid
b. Carnosine
c. Homocarnosine
d. Anserine
Solution. Ans-150 : (c) Homocarnosine
Ref.: Read the text below
Sol :
- Homocarnosine is a dipeptide composed of GABA and histidine.
All other 3 compounds contain β–Alannine. Beta alanine
- Only naturally occurring β–amino acid
- Formed from catabolism of pyrimidines – cytosine and uracil
- Required for synthesis of pan
Correct Answer. c
(151).
Obermeyer test is positive in which of the following diseases?
a. Tyrosinemia
b. Hartnup disease
c. Maple syrup urine disease
d. Alkaptanuria
Solution. Ans-151 : (b) Hartnup disease
Ref.: Read the text below
Sol :
Obermeyer Test :
- It is a test to detect indicans in urine
- It is positive in Hartnup’s disease in which there is defective absorption of tryptophan in intestine and renal tubules
- The test is an indicator of intestinal toxemia and overgrowth of anaerobic bacteria
- Indican is an indole produced when bacteria in the intestine act on the amino acid, tryptophan
- Most indoles are excreted in the feces. The remainder is absorbed, metabolized by the liver, and excreted as indicant in the
urine.
Correct Answer. b
(152).
All of the following are required in PCR except?
a. Deoxyribonucleotides
b. Thermostable enzyme
c. Dideoxyribonucleotides
d. DNA Template
Solution. Ans-152 : (c) Dideoxyribonucleotides
Ref.: Read the text below
Sol : A basic PCR set up requires several components and reagents. These components include:
-DNA template that contains the DNA region (target) to be amplified.
- Two primers that are complementary to the 3' (three prime) ends of each of thesense and anti-sense strand of the DNA target.
- Taq polymerase or another DNA polymerase with a temperature optimum at around 70 °C. - Deoxynucleoside triphosphates (dNTPs,
sometimes called "deoxynucleotide triphosphates"; nucleotides containing triphosphate groups), the building-blocks from which the DNA
polymerase synthesizes a new DNA strand.
- Buffer solution, providing a suitable chemical environment for optimum activity and stability of the DNA polymerase.
- Divalent cations, magnesium or manganese ions; generally Mg2+ is used, but Mn2+ can be utilized for PCR-mediated DNA
mutagenesis, as higher Mn2+concentration increases the error rate during DNA synthesis[9]
- Monovalent cation potassium ions.
Correct Answer. c
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(153).
Subtelomeric rearrangement of genes is frequently associated with mental retardation. All of the following techniques can be used to
diagnose them except?
a. FISH
b. MAPH
c. CGH array
d. MALDI
Solution. Ans-153 : (d) MALDI
Ref.: Read the text below
Sol :
- It is a technique used to assess the molecule, based on their mass only.
- Most commonly employed methods for dispersing peptides, proteins, and other large biomolecules into the vapor phase for mass
spectrometric analysis are electrospray ionization and matrix – assisted laser desorption and ionization, or MALDI.
- In electrospray ionization, the molecules to be analyzed are dissolved in a volatile solvent and introduced into the sample chamber in a
minute stream through a capillary. As the droplet of liquid emerges into the sample chamber, the solvent rapidly disperses leaving the
macromolecule suspended in the gaseous phase. The charged probe serves to ionize the sample.
- Electrospray ionization is frequently used to analyze peptides and proteins as they elute from an HPLC or other chromatography
column.
Correct Answer. d
(154).
If more than one codon codes for same amino acid, this phenomenon is known as?
a. Degeneracy
b. Frame – Shift mutation
c. Transcription
d. Mutation
Solution. Ans-154 : (a) Degeneracy
Ref.: Read the text below
Sol :
- Three of the 64 possible codons do not code for specific amino acids; these have been termed nonsense codons. These nonsense codons
are utilized in the cell as termination signals; they specify where the polymerization of amino acids into a protein molecule is to stop. The
remaining 61 codons code for 20 amino acids. Thus, there must be “degeneracy” in the genetic code – i.e. multiple codons must decode
the same amino acid. Some amino acids are encoded by several codons, e.g. six different codons specify serine
- Other amino acids, such as methionine and tryptophan, have a single codon.
- In general, the third nucleotide in a codon is less important than the first two in determining the specific amino acid to be
incorporated, and this accounts for most of the degeneracy of the code.
Correct Answer. a
(155).
Chymotrypsinogen is a ?
a. Zymogen
b. Carboxypeptidase
c. Transaminase
d. Clot lysing protein
Solution. Ans-155 : (a) Zymogen
Ref.: Read the text below
Sol :
- The proteases are secreted as inactive zymogens; the active site of the enzyme is masked by a small region of the peptide chain that is
removed by hydrolysis of a specific peptide bond.
- Pepsinogen is activated to pepsin by gastric acid and by activated pepsin (autocatalysis).
- In the small intestine, trypsinogen, the precursor of trypsin, is activated by enteropeptidase, which is secreted by the duodenal
epithelial cells; trypsin can then activated chymotrypsinogen to chymotrypsin, progelastase to eleastase, procarboxy-peptidase to
carboxypeptidase, and proaminopeptidase to aminopeptidase.
Correct Answer. a
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(156).
Sommer’s movements in Forensic Medicine are associated with?
a. Decomposition
b. Rigor mortis
c. Post mortem lividity
d. Mummification
Solution. Ans-156 : (b) Rigor mortis
Ref.: Read the text below
Sol :
- The view that the development of rigor mortis could produce significant movements of the body was promoted by Sommer, in about
1833, and the postulated movements became known as "Sommer's movements".
- It is now accepted that movements of a corpse due to the development of rigor mortis can only occur in special circumstances, such as
an extreme position of the body at the moment of death.
- If a body is moved before the onset of rigor then the joints will become fixed in the new position in which the body is placed. For this
reason, when a body is found in a certain position with rigor mortis fully developed, it cannot be assumed that the deceased necessarily
died in that position.
- Conversely, if the body is maintained by rigor in a position not obviously associated with support of the body, then it can be concluded
that the body was moved after rigor mortis had developed.
Correct Answer. b
(157).
Which amongst the following is a law associated with cadaveric rigidity?
a. Durham’s rule
b. Curren’s rule
c. Nysten’s rule
d. Nettler’ rule
Solution. Ans-157 : (c) Nysten’s rule
Ref.: Read the text below
Sol :
- Ordinarily, death is followed immediately by total muscular relaxation - primary muscular flaccidity - succeeded in turn by generalised
muscular stiffening - rigor mortis. After a variable period of time rigor mortis passes off spontaneously to be followed by secondary
muscular flaccidity. The first investigation of rigor mortis is attributed to Nysten in 1811.
- Nysten's law rigor mortis affects first the muscles of mastication, next those of the face and neck, then those of the trunk and arms, and
last those of the legs and feet.
Correct Answer. c
(158).
“Smoking Stool Syndrome” is seen in?
a. Yellow phosphorus
b. Red phosphorus
c. Both
d. None of the above
Solution. Ans-158 : (a) Yellow phosphorus
Ref.: Read the text below
Sol :
- The accepted lethal dose when white phosphorus is ingested orally is 1 mg per kg of body weight, although the ingestion of as little as
15 mg has resulted in death.
- It may also cause liver, heart or kidney damage.
- There are reports of individuals with a history of oral ingestion who have passed phosphorus-laden stool ("smoking stool syndrome").
- Its extreme toxicity is due to the generation of free radicals, especially in the liver, where they accumulate and are not easily
metabolized.
Correct Answer. a
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(159).
Father of Modern Toxicology is?
a. J. Cameron
b. Alfred
c. Orfila
d. Rokitansky
Solution. Ans-159: (c) Orfila
Ref.: Read the text below
Sol :
- Mathieu Orfila is considered to be the modern father of toxicology, having given the subject its first formal
treatment in 1813 in his Traité des poisons, also called Toxicologie générale.
Correct Answer. c
(160).
Kunkel’s test is done to demonstrate the presence of
a. Lead
b. CuS04
c. CO
d. Dhatura
Solution. Ans-160 : (c) CO
Ref.: Read the text below
Sol :
Correct Answer. c
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(161).
“Ewing’s postulates” refer to?
a. Accidents as a cause by birth
b. Complication resulting from trauma
c. Circulation, respiration and brain functions
d. Circulation, respiration and excretion
Solution. Ans-161 : (b) Complication resulting from trauma
Ref.: Read the text below
Sol :
- Ewing's postulate showing relationship between trauma and new growth
- The pathophysiology of burn scar neoplasms is clearly distinct from that of regular skin cancer. Ewing's
postulate requires: (1) evidence of a burn scar; (2) a tumor within the boundaries of the scar; (3) no previous
tumor in that location; (4) tumor histology that is compatible with the cell types found in the skin and the scar; and
(5) an adequate interval between the burn injury and tumor development
Correct Answer. b
(162).
Narcotic Drugs and Psychotropic substances act was passed in the year?
a. 1981
b. 1983
c. 1985
d. 1986
Solution. Ans-162 : (c) 1985
Ref.: Read the text below
Sol :
THE NARCOTIC DRUGS AND PSYCHOTROPIC SUBSTANCES ACT,1985
- An Act to consolidate and amend the law relating to narcotic drugs, to make stringent
provisions for thecontrol and regulation of operations relating to narcotic drugs and psychotropic substances 1[, to provide forthe
forfeiture of property derived from, or used in, illicit traffic in narcotic drugs and psychotropic substances, toimplement the provisions of
the International Convention on Narcotic Drugs and Psyc hotropic Substances]and for matters connected therewith
Correct Answer. c
(163).
The process which causes drying up of tissues and internal viscera to a sufficient degree to halt putrefaction is called?
a. Saponification
b. Adipocere formation
c. Mummification
d. Putrefaction
Solution. Ans-163 : (c) Mummification
Ref.: Read the text below
Sol :
- The process which causes drying up of tissues and internal viscera to a sufficient degree to halt putrefaction is called mummification.
- Mummification occurs in hot dry environments where the body can dehydrate quickly and bacterial action is minimised.
- The process can take place in days or weeks and the skin would appear dark, leathery and dry while the internal organs would
desiccate and reduce in size
Correct Answer. c
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(164).
Temperature of the body rises up for the first 2 hrs after death. The probable condition includes the following except
a. Sun stroke
b. Frost bite
c. Septicemia
d. Tetanus
Solution. Ans-164 : (b) Frost bite
Ref.: Read the text below
Sol : Postmortem caloricity-temperature raised for first 2 hrs
- Sunstroke
- In some nerve disorders
- Tetanus
- Strychnine poisoning
- Excessive bacterial activity in septicemia
- Cholera
Correct Answer. b
(165).
Which of the following is known as ‘road poison’?
a. Cannabis
b. Chloroform
c. Dhatura
d. Aconite
Solution. Ans-165: (c) Dhatura
Ref.: Read the text below
Sol :
- Datura is also known as 'Road poison'.
- All Datura plants contain tropane alkaloids such as scopolamine, hyoscyamine, and atropine, primarily in their seeds and flowers.
Because of the presence of these substances, Datura has been used for centuries in some cultures as a poison and as a hallucinogen.
- There can be a 5:1 toxin variation across plants, and a given plant's toxicity depends on its age, where it is growing, and the local
weather conditions.
- This variation makes Datura exceptionally hazardous as a drug.
Correct Answer. c
(166).
Phossy jaw is seen in which poisoning ?
a. Mercury
b. Yellow phosphorous
c. Red phosphorus
d. Tetanus
Solution. Ans-166 : (c) Red phosphorus
Ref: Read the text below
Sol:
- Phossy jaw is seen in chronic poisoning with the fumes of Red phosphorus, seen in the industries where its use is rampant like
matchbox industry and incense making..
Correct Answer. c
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(167).
RDX is one of the most important and useful military explosives. Its full form is :
a. Real death X-ray
b. Red dye xylophone
c. Rightful death explosive
d. Research Department Explosive
Solution. Ans-167 : (d) Research Department Explosive
Ref.: Read the text below
Sol :
- RDX, an initialism for Research Department Explosive, is an explosive nitroamine widely used in military and industrial applications. It
was developed as an explosive which was more powerful than TNT, and it saw wide use in WWII. RDX is also known as cyclonite,
hexogen (particularly in German and German-influenced languages), and T4.
- Its chemical name is cyclotrimethylenetrinitramine; name variants include cyclotrimethylene-trinitramine and cyclotrimethylene
trinitramine.
Correct Answer. d
(168).
Typical injury in a bomb explosion is :
a. Abrasion
b. Punctate bruises
c. Puncture laceration
d. A triad of punctuate bruises, abrasions and puncture lacerations
Solution. Ans-168 : (d) A triad of punctuate bruises, abrasions and puncture lacerations
Ref.: Read the text below
Sol : Typical injury in a bomb explosion is triad of punctuate bruises, abrasions and puncture lacerations
Correct Answer. d
(169).
Arborescent markings are due to :
a. Dry burns
b. Scalds
c. Hanging
d. Lightning
Solution. Ans-169 : (d) Lightning
Ref.: Read the text below
Sol :
- These markings are known by several different names, some among them being “arborization” , “feathering”, “ferning”, “filigree
burns”, “arborescent burns”, “:Lichtenberg’s flowers” *after a preeminent German physicist and scientist Georg christoph Lichtenberg
who reproduced similar images with the help of static electricity]
Correct Answer. d
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(170).
Which of the following is a characteristic postmortem finding of a death due to electrocution ?
a. Honeycomb liver
b. Local hemorrhages and/or tears in the muscles
c. Tache’ noir de la sclerotiques
d. All of the above
Solution. Ans-170 : (b) Local hemorrhages and/or tears in the muscles
Ref.: Read the text below
Sol :
- The hemorrhages/tears are especially abdundant in muscles of the extremity. These are due to forcible contraction and spasm induced
by the passage of current.
Correct Answer. b
(171).
Counter-coup injury is seen in :
a. Gunshot wound
b. Head injury
c. Wound in the abdomen
d. Chest wounds
Solution. Ans-171 : (b) Head injury
Ref.: Read the text below
Sol : A variant of this question is : Counter-coup injuries are seen in
- Brain
- Heart
- Liver
- Pancreas
Correct Answer. b
(172).
The most common type of haemorrhage seen in boxers is :
a. Extradural
b. Subdural
c. Subarachnoid
d. Intracerebral
Solution. Ans-172 : (b) Subdural Ref.: Read the text below Sol :
- Extradural haemorrhage almost never occurs in boxers, since fractures of skull occur rarely.
- Subarachnoid haemorrhage may occur rarely if berry aneurysm was present.
Correct Answer. b
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(173).
Which of the following is more dangerous in causing spinal damage ?
a. Hyperextension of the neck
b. Hyperflexion of the neck
c. Both hyperflexion and hyperextension of the neck
d. None can cause spinal damage
Solution. Ans-173 : (a) Hyperextension of the neck Ref.: Read the text below Sol :
- Hyperextension is more dangerous because flexion is protected by the contraction of strong posterior neck muscles.
- In comparison, the anterior longitudinal ligament (which would protect from injuries during hyperextension) is quite weak.
Correct Answer. a
(174).
Greenish color in a contusion is due to :
a. Haemosiderin
b. Haematoidin
c. Sulphmethemoglobin
d. Methaemoglobin
Solution. Ans-174 : (b) Haematoidin
Ref.: Read the text below
Sol :
- Haematoidin is a loose term formerly applied to the crystals of bilirubin or biliverdin.
- Besides occurring in old contusions it also sometimes occurs in faeces after intestinal haemorrhage.
- When present in the corpora lutea it is called haemolutein. Important thing to remember is that while haemosiderin contains iron,
haematoidin is iron free.
Correct Answer. b
(175).
In a large bruise, color changes begin :
a. From the center
b. From the periphery
c. From an area somewhere in the middle of centre and periphery
d. At the same time everywhere
Solution. Ans-175 : (b) From the periphery
Ref.:Read the text below
Sol :
- A large bruise takes longer time to heal than a smaller bruise. The color changes begin at the periphery first.
A large old bruise may contain all possible colors seen in a bruise – from purple in the centre to yellow at the edges.
Correct Answer. b
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(176).
A motor car running over the thigh, leaving tire marks over it is a good example of :Ectopic bruise
a. Ectopic bruise
b. Fictitious bruise
c. Gravitating bruise
d. Patterned bruise
Solution. Ans-176 : (d) Patterned bruise Ref.: Read the text below Sol :
- Patterned bruise are important from a medicolegal point of view, because they can give a fairly good idea of the striking surface.
- Among the most commonly encountered patterned injuries are patterned abrasions, patterned bruises and patterned lacerations.
- These are so called because these injuries reflect the pattern of the striking surface.
- Beating with whips having patterned thongs and kicking with boots may leave patterned.
Correct Answer. d
(177).
Gutter fracture is caused by :
a. A firearm missile glancing along the outer table of the skull
b. Falling into a shallow gutter
c. Falling into a river or any other body of water
d. Both (b) and (c)
Solution. Ans-177: (a) A firearm missile glancing along the outer table of the skull
Ref.: Read the text below
Sol :
- These are called so, because the firearm missile glancing along the outer table would create a furrow, which looks like a gutter.
- This fracture is frequently associated with comminuted depressed fractures of the inner table of the skull.
Correct Answer. a
(178).
A fracture in which a piece of the bone is removed by violence, as by a bullet is known as :
a. Resecting fracture
b. Secondary fracture
c. Segmental fracture
d. Pressure fracture
Solution. Ans-178: (a) Resecting fracture
Ref.: Read the text below
Sol :
- Secondary fracture – also known as spontaneous fracture of pathologic fracture – is caused due to weakening of the bone structure by
pathologic processes, such as neoplasia, osteomalacia, osteomyelitis, and other diseases.
- Segmental fracture, also known as double fracture is the fracture of a bone in two places.
- Pressure fracture is caused by pressure on the bone from an adjoining tumor.
Correct Answer. a
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(179).
Essential pentosuria is due to defect in ?
a. HMP shunt pathway
b. Glycolysis
c. Gluconeogenesis
d. Uronic acid pathway
Solution. Ans-179: (d) Uronic acid pathway Ref: Read the text below Sol :
- Essential pentosuria is due to deficiency of enzyme converting L-xylulose to xylitol using reducing equivalent from NADPH.
- Also remember : vitamin C also gets synthesized in the uronic acid pathway.
- In humans and other primates, as well as in guinea pigs, bats, some birds and fishes, ascorbic acid cannot be synthesized because of the
absence of L-gulonolactone oxidase.
Correct Answer. d
(180).
Factor common for both glycolysis and gluconeogenesis is ?
a. Phosphofructokinase
b. Fructose 2, 6 bisphosphatase
c. Hexokinase
d. Glucose 6 phosphatase
Solution. Ans-180 : (b) Fructose 2, 6 bisphosphatase Ref: Read the text below Sol :
- Phospofructokinase and hexokinase are the enzyme important in glycolysis alone, while glucose 6 phosphatase is the enzyme which is
utilized in gluconeogenesis.
- Glucose 6 phosphatase is also used in Glycogenolysis in the liver.
- This enzyme is not used in Glycolysis so the answer of above question is Fructose 2, 6 bisphosphate as it has got effect both on PFK-1
and on Frucose 1, 6 bisphosphates enzyme.
Correct Answer. b
(181).
Fructokinase reaction produces which of the following intermediates ?
a. Fructose – 1 – phosphate
b. Fructose – 6 – phosphate
c. Fructose 1, 6 diphosphate
d. Glyceraldehydes and dihydroxyacetone phosphate
Solution. Ans-181 : (a) Fructose – 1 – phosphate Ref: Read the text below Sol :
- First step of fructose metabolism is catalysed by fructokinase and the product is fructose-1-phosphate.
Correct Answer. a
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(182).
Which of the following enzyme does not catalyse decarboxylation reaction ?
a. Pyruvate dehydrogenase
b. α–ketoglutarate dehydrogenase
c. Phosphoenol pyruvate carboxykinase
d. Pyruvate carboxylase
Solution. Ans-182 : (d) Pyruvate carboxylase
Ref:Read the text below
Sol :
- Pyruvate carboxylase convertspyruvate to oxaloacetate by incorporation of CO2.
- So it is a carboxylation reaction and not the decarboxylation reaction.
- PHD, α KG Dehydrogenase and Phosphoenol pyruvate carboxykinase catalyse decarboxylation reaction.
Correct Answer. d
(183).
Which enzyme does not act in fructose metabolism ?
a. Fructokinase
b. Aldolase A
c. Aldolase B
d. Pyruvate kinase
Solution. Ans-183 : (b) Aldolase A
Ref: Read the text below
Sol :
- Only Aldolase B which is found exclusively in liver acts in fructose metabolism.
- Aldolase which is universal in localization acts on glycolysis.
- (In glycolyis, Aldolase A or Aldolase B may be used for splitting fructose 1,6 bisphosphate to DHAP and Glyceraldehyde 3 phosphate).
Correct Answer. b
(184).
In the skull, fractures caused not at the point of impact, but some distance away from it are known as :
a. Diagonal fractures
b. Atypical skull fractures
c. Bursting fractures
d. Depressed fractures
Solution. Ans-184 : (c) Bursting fractures
Ref.: Read the text below
Sol :
- Bursting fractures of the skull are often seen in falls from heights, when the vertex undergoes sudden flattening.
- They are also seen in contact firearm wounds of the skull, where the fractures are often seen away from where the bullet enters the
skull.
Correct Answer. c
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(185).
Which of the following signs would indicate that the death was antemortem ?
a. Cutis anserine
b. Washer woman’s hands
c. Weed and grass in the hands
d. Cyanosis
Solution. Ans-185 : (c) Weed and grass in the hands
Ref.: Read the text below
Sol :
- Cutis anserine and washer woman’s hands are not signs of antemortem drowing.
- Cutis anserina may be seen in any death due to rigor mortis of erector pilae muscles.
- Washer woman’s hands are merely signs of submersion.
Correct Answer. c
(186).
Oligohalophilic diatoms live in :
a. Fresh water
b. Brackish water
c. Sea water
d. Sewer water
Solution. Ans-186 : (a) Fresh water
Ref.: Read the text below
Sol :
- Oligohalophilic diatoms live in fresh water with a salinity of less than 0.05%.
- Polyhalophilic diatoms live in sea water.
- Mesohalophilic diatoms live in water midway in salinity between freshwater and seawater, i.e. brackish water.
Correct Answer. a
(187).
For diatom test, the best site for taking samples is :
a. Lungs
b. Bone marrow in ulna
c. Bone marrow in femur
d. A large and fleshy muscle
Solution. Ans-187 : (c) Bone marrow in femur
Ref.: Read the text below
Sol :
- Sometimes the question is asked in this form – Antemortem drowing can best be confirmed by demonstrating diatoms in (a) Brain and
bone marrow (b) Liver and anal canal (c) Pancreas and lungs (d) Lungs and stomach.
- The answer will be brain and bone marrow.
Correct Answer. c
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(188).
Emphysema aquosum is seen in :
a. Wet drowning
b. Dry drowning
c. Secondary drowning
d. Immersion syndrome
Solution. Ans-188 : (a) Wet drowing
Ref.: Read the text below
Sol :
- The term “Emphysema aquosum” (also known as emphysema hydroaerique) and “Oedema aquosum” are rarely used, if at all, by serious
forensic pathologists today.
- The only justification of this question being included here is that this question is still being asked in various examinations.
Correct Answer. a
(189).
Subtelomeric rearrangement of genes is frequently associated with mental retardation. All of the following techniques can be used to
diagnose them except?
a. FISH
b. MPH
c. cGH array
d. MALDI
Solution. Ans-189 : (d) MALDI Ref.: Read the text below Sol :
- It is a technique used to assess the molecule based on their mass only.
- Most commonly employed methods for dispersing peptides, proteins, and other large biomolecules into the vapor phase for mass
spectrometric analysis are electrospray ionization and matrix-assisted laser desorption and ionization, or MALDI.
Correct Answer. d
(190).
Chromosomal mutations can be identified from all except.
a. Single stranded conformational polymorphism
b. Dideoxy nucleotide trail sequencing
c. Agarosge electrophoresis
d. Denaturing Gradient Gel Electrophoresis
Solution. Ans-190 : (d) Denaturing Gradient Gel Electrophoresis Ref.: Read the text below Sol :
- Gradient gel electrophoresis is used, though not widely for separation of individual lipoprotein. It has got nothing to do with finding out
the mutation.
- Rest all other method given in the options are used for detection of the mutation.
Single stranded conformational polymorphism (SSCP)
- If the base change is not changing the restriction site, that time RFLP can not be done to detect the base change, that time SSCP could
be done to find out the base change.
- Basic principal of SSCP is that movement of smaller DNA fragment on the electrophoresis is partially dependent on their conformation,
a single base change usually modified the conformation of the DNA sequence sufficiently to cause the mobility shift on the
electrophoresis through a PAGE.
Correct Answer. d
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(191).
To produce insulin for therapeutic purpose in bacteria involves which of the following as initial material?
a. m RNA from beta pancreatic cell of human
b. Genomic DNA from beta pancreatic cell of human
c. m RNA from lymphocyte of human
d. Genomic cDNA from lymphocyte of human
Solution. Ans-191 : (a) m RNA from beta pancreatic cell of human Ref.: Read the text below Sol :
- Restriction fragment of the DNA of an organism is collected in DNA library after recombinant DNA technology.
- Joining of two different DNA via DNA ligase enzyme is known as recombinant DNA technology.
- Depending upon the source of the DNA used to insert in to the vector DNA for recombinant DNNA technology, the gene library is of two
types.
Correct Answer. a
(192).
What is the age at which a woman can give a valid consent for MTP under the MTP Act 1971 ?
a. 12 years
b. 16 years
c. 18 years
d. No age has been defined
Solution. Ans-192 : (c) 18 years Ref.: Read the text below Sol :
- Three ages that are very important from a medicolegal angle, and which are quite often confused together are 12 years, 16 years and
18 years. Please remember the following facts :
- 12 years : Age of consent for general physical examination.
- 16 years : Age of consent for a girl for sexual intercourse.
- 18 years : Age of consent for MTP.
Correct Answer. c
(193).
“Utus Paste” was once commonly used to procure criminal abortion. Which was its main component responsible for causing abortions ?
a. Iodine or some salt of iodine such as potassium iodide
b. Oil of pennyroyal
c. Borax
d. Some salt of lead
Solution. Ans-193 : (a) Iodine or some salt of iodine such as potassium iodide
Ref.: Read the text below
Sol :
- “Utus paste” is a mixture of soap, myrrh resinoid and potassium iodide, where the iodide acted as an irritant causing abortion.
- It was once used for legitimate abortions as well, but soon became popular with criminal abortionists.
- It was particularly useful in late abortions. This paste is injected into the cervix from a collapsible tube (much like a modern toothpaste
tube) with a uterine applicator.
Correct Answer. a
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(194).
Disputed paternity or maternity can be solved by using all the following except :
a. Blood grouping
b. HLA typing
c. Precipitin test
d. DNA profiling
Solution. Ans-194 : (c) Precipitin test
Ref.: Read the text below
Sol :
- Precipitin test is used to determine the species origin of blood. All other tests can be used to solve disputed paternity or maternity.
- With DNA profiling, it can be done in 100% cases, but with blood grouping and HLA typing it is useful in select cases.
Correct Answer. c
(195).
Takayama reagent is used in :
a. Guaiacum test
b. Kastle-Meyer test
c. Haemin crystal test
d. Haemochromogen crystal test
Solution. Ans-195 : (d) Haemochromogen crystal test
Ref.: Read the text below
Sol :
- Takayama reagent is used in haemochromogen crystal test. The crystals are of salmon pink color. Shape is feathery.
- No heating is required.
- Slide is examined under the microscope just in five minutes. Test is of more value than Teichman test or Haemin crystal test.
Correct Answer. d
(196).
Which of the following tests can best be used to determine the human origin of a blood stain ?
a. Precipitin test
b. Benzidine test
c. Haemin crystal test
d. Any of the above
Solution. Ans-196 : (a) Precipitin test
Ref.: Read the text below
Sol :
- Benzidine test is merely used for screening of blood. It does not confirm blood.
- Haemin crystal test confirms blood, but does not give the species.
- It is the precipitin test which gives the species.
Correct Answer. a
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(197).
Substrate level phosphorylation in TCA cycle occurs at?
a. Succinyl Co A to succinate
b. Fumarate to malate
c. Succinate to fumarate
d. Acetoacetate to a KG
Solution. Ans-197 : (a) Succinyl Co A to succinate
Ref.: Read the text below
Sol :
- In TCA cycle there occurs substrate level phosphorylation only once at the step catalysed by succinate thiokinase which produces ATP
in most of the cells.
- GTP at this step is produced in few of the cycle in gluconeogenic organs, i.e. liver and kidney.
Correct Answer. a
(198).
All the following TCA cycle intermediate may be added or removed by other metabolic pathways except?
a. Citrate
b. Fumarate
c. Isocitrate
d. Alpha ketoglutarate
e. Oxaloacetate
Solution. Ans-198 : (c) Isocitrate
Ref.: Read the text below
Sol :
- Citrate is transported out of the mitochondria to be used as a source of cytoplasmic acetyl Co A.
- Fumarate is produced during degradation of tyrosine and phenylalanine.
- Alpha ketoglutarate can be formed from glutamate.
- Oxaloacetate may be produced from pyruvate during gluconeogenesis.
Correct Answer. c
(199).
Composition of hyaluronic acid?
a. N-acetyl glucosamine + d-glucosamine acid
b. N-acetyl glucosamine + d-glucuronic acid
c. N-acetyl glucosamine + sulfated glucosamine acid
d. N-acetyl glucosamine + iduranic acid.
Solution. Ans-199 : (b) N-acetyl glucosamine + d-glucuronic acid
Ref.: Read the text below
Sol :
- Hyaluronan is a polymer of disaccharides, themselves composed of D-glucuronic acid and D-N-acetylglucosamine, linked via alternating
β-1,4 and β-1,3 glycosidic bonds.
- Hyaluronan can be 25,000 disaccharide repeats in length. Polymers of hyaluronan can range in size from 5,000 to 20,000,000 Da in
vivo.
- The average molecular weight in human synovial fluid is 3−4 million Da, and hyaluronan purified from human umbilical cord is
3,140,000 Da
Correct Answer. b
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(200).
Phospholipid include all except?
a. Plasmalogen
b. Dipalmitoyl lecithin
c. Ceramide
d. Cardiolipin
Solution. Ans-200: (c) Ceramide
Ref.: Read the text below
Sol :
- Ceramide is sphingosine and fatty acid. It is not a phospholipid. Dipalmitoyl lecithin and Cardiolipin are phospholipids.
- Plasmalogens : Plasmalogen are also a special type of phospholipid. These compounds constitute as much as 10% of the phospholipids
of brain and muscle. Structurally, the plasmalogens resemble phosphatidylethanolamine but possess an ether link on the sn-1 carbon
instead of the ester link found in acylglycerols.
- Typically, the alkyl radical is an unsaturated alcohol. In some instances, choline, serine, or inositol may be substituted for ethanolamine.
Correct Answer. c
Test Answer
1.(a)
2.(c)
3.(c)
4.(c)
5.(b)
6.(d)
7.(a)
8.(d)
9.(b)
10.(d)
11.(b)
12.(a)
13.(a)
14.(b)
15.(a)
16.(b)
17.(d)
18.(a)
19.(b)
20.(a)
21.(c)
22.(b)
23.(d)
24.(b)
25.(d)
26.(d)
27.(c)
28.(a)
29.(a)
30.(a)
31.(c)
32.(d)
33.(a)
34.(b)
35.(a)
36.(d)
37.(b)
38.(d)
39.(d)
40.(c)
41.(a)
42.(d)
43.(a)
44.(d)
45.(c)
46.(c)
47.(d)
48.(a)
49.(d)
50.(c)
51.(d)
52.(d)
53.(c)
54.(d)
55.(d)
56.(d)
57.(a)
58.(d)
59.(d)
60.(c)
61.(d)
62.(d)
63.(d)
64.(a)
65.(c)
66.(d)
67.(c)
68.(a)
69.(a)
70.(d)
71.(c)
72.(d)
73.(d)
74.(b)
75.(c)
76.(b)
77.(a)
78.(a)
79.(c)
80.(c)
81.(b)
82.(c)
83.(a)
84.(a)
85.(d)
86.(b)
87.(c)
88.(a)
89.(b)
90.(c)
91.(a)
92.(b)
93.(d)
94.(d)
95.(a)
96.(b)
97.(d)
98.(c)
99.(d)
100.(d)
101.(c)
102.(b)
103.(d)
104.(b)
105.(c)
106.(a)
107.(b)
108.(b)
109.(c)
110.(c)
111.(a)
112.(d)
113.(a)
114.(c)
115.(a)
116.(d)
117.(a)
118.(c)
119.(c)
120.(a)
121.(c)
122.(a)
123.(a)
124.(d)
125.(a)
126.(c)
127.(d)
128.(b)
129.(d)
130.(b)
131.(b)
132.(c)
133.(c)
134.(b)
135.(c)
136.(c)
137.(b)
138.(a)
139.(d)
140.(d)
141.(a)
142.(a)
143.(a)
144.(b)
145.(a)
146.(d)
147.(b)
148.(d)
149.(b)
150.(c)
151.(b)
152.(c)
153.(d)
154.(a)
155.(a)
156.(b)
157.(c)
158.(a)
159.(c)
160.(c)
161.(b)
162.(c)
163.(c)
164.(b)
165.(c)
166.(c)
167.(d)
168.(d)
169.(d)
170.(b)
171.(b)
172.(b)
173.(a)
174.(b)
175.(b)
176.(d)
177.(a)
178.(a)
179.(d)
180.(b)
181.(a)
182.(d)
183.(b)
184.(c)
185.(c)
186.(a)
187.(c)
188.(a)
189.(d)
190.(d)
191.(a)
192.(c)
193.(a)
194.(c)
195.(d)
196.(a)
197.(a)
198.(c)
199.(b)
200.(c)
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