IIT JAM Biology 2008
Previous Year Question Paper with Solution.

1. In cells, cellulose and glycogen function as

    (a) energy storage components

    (b) structural and energy storage components, respectively

    (c) energy storage and structural components, respectively

    (d) structural components

Ans.    (b)

Sol.    Cellulose is basically a polysaccharide consisting of tinear chain of several units linked together by (1 → 4) linked D-glucose. It is a structural component of the primary cell wall of green plants and many algae. Where as glycogen is a polysaccharides of glucose which function as energy storage in animals cells (i.e. in liver and muscles). It is also called as animal starch.

2. The structures of myoglobin and hemoglobin are

    (a) Quaternary

    (b) Quaternary and teritary, respectively

    (c) Tertiary and quaternary, respectively

    (d) Molten globule

Ans.    (c)

Sol.    Myoglobin is a monomeric hemoprotein found mainly in muscle tissue where it serves as an intracellular storage site for O2. It has a tertiary structure like that of a typical water soluble globular protein.

    Haemoglobin is an , tetrameric hemeprotein found in erythrocytes where it is responsible for binding oxygen in the lung and transporting the bound oxygen throughout the body where it is used in aerobic metabolic pathways.

3. The chromatographic technique for determination of the native molecular weight of proteins is

    (a) Gel permeation

    (b) Ion exchange

    (c) Hydrophobic interaction

    (d) Chromatofocusing

Ans.    (a)

Sol.    Gel permentation chromatography (GPC) or size exclusion chromatography is a technique of separation of molecules based on their hydrodynamic volume. This technique is widely used for polymer molecular weight determination. Thus, we can use chromatography technique for determination of the native molecular weight of proteins.

4. The affinity of an enzyme to the substrate is indicated by

    (a) pKa

    (b) Ki

    (c) kcat

    (d) Km

Ans.    (d)

Sol.    Km is the substrate concentration at which initial velocity becomes half of its maximum velocity used in enzyme kinetics.

    Thus, to denote affinity of an enzyme to the substrate we use Km. It is expressed in molar units.

5. Which one of the following compounds is optically INACTIVE?

    (a) Ala

    (b) Cys

    (c) Gly

    (d) Lys

Ans.    (c)

Sol.    Optical activity can be quantified by the rotation of the plane of polarised light as it passes through a substance. Organic compounds containing chiral center show optical activity. Thus, we know that except glycine, all other three, i.e. alanine, cystein and lysine have chiral carbons in their structure. Therefore, only glycine would not show optical activity.

6. In O-linked glycoproteins, the glycan part is linked to the polypeptide. The amino acid residues involved in this linkage are

    (a) Ser and Thr

    (b) Gln and Tyr

    (c) Tyr and Thr

    (d) Asn and Tyr

Ans.    (a)

Sol.    O-linked glycosylation is the attachment of a sugar molecule to an oxygen atom in an amino acid residue in a protein. It occurs in the Golgi apparatus in eukaryotes during later stages of protein processing. It is the addition of N-acetyl galactosamine to serine or threonine residues by the enzyme UDP-N-acetyl-D-galactosamine.

7. The cofactor (s) required for the nitrogenase Qenzyme complex involved in N2 fixation is/are

    (a) Fe and Mo

    (b) Fe

    (c) Fe and S

    (d) Fe, S and Mo

Ans.    (d)

Sol.    Nitrogenase enzyme system catalyses the reduction of molecular nitrogen to ammonia during biological nitrogen fixation. Cofactor required for the nitrogenase enzyme complex is molybdenum nitrogenase consisting of two components : nitrogenase molybdenum iron (M0Fe) protein or dinitrogenase and nitrogenase iron (Fe) protein or nitrogenase reductase.

    The M0Fe protein is an tetramer. Nitrogenase Fe protein is a dimer of two identical subunits that coordinate a sing [Fe4S4] cluster.

8. The anaplerotic (filling up) reaction to replenish citric acid cycle is

    (a) decarboxylation of isocitrat to α-ketoglutarate

    (b) decarboxylation of α-ketoglutarate of succinyl CoA

    (c) carboxylation of phosphoenolpyruvate to oxaloacetate

    (d) 1, 2 and 1

Ans.    (c)

Sol.    Anaplerosis is the act of replenishing TCA cycle (citric and cycle) intermediate that have been extracted for biosynthesis. The TCA cycle is a hub of metabolism, with central importance in both energy production and biosynthesis. One of the most important anaplerotic reaction is the production of oxaloacetate from pyruvate.


9. Which one of the following is NOT a neurotransmitter?

    (a) Glutamine

    (b) Glutamate

    (c) Glycine

    (d) Acetycholine

Ans.    (a)

Sol.    Glutamine is not a neurotransmitter, infact it is an essential and most abundant amino acid present in our body.

    Neurotransmitters are endogenous chemicals that transmits signals across a synapse or junction from one neuron (nerve cell) to another target neuron, muscle cell or gland cell.

10. Which one of the following is NOT an energy rich compound?

    (a) Phosphoenolpyruvate

    (b) Glucosel, 6-bisphosphate

    (c) Acetyl phosphate

    (d) Phosphoarginine

Ans.    (b)

Sol.    Energy rich compounds are those with a standard free energy of more than 6 kcal/mol.

    These are formed by the oxidation of substrate which cell obtains from the environment. Energy rich compounds in cells comprise of five kinds of high energy bonds as follows.


11. During photosynthesis, ATP synthesis takes place in

    (a) stroma

    (b) thylacoid lumen

    (c) thylacoid membrane

    (d) cytoplsm

Ans.    (c)

Sol.    In plants, ATP is synthesised in thylacoid membrane of the chloroplast during the light dependent reactions of photosynthesis, i.e. photophosphorylation. Here light energy is used to pump protons across the chloroplast membrane. This produces a proton-motive force and this drives the ATP synthase, exactly as in oxidative phosphorylation.

12. Contact inhibition phenomenon is observed in

    (a) animal cell culture

    (b) plant cell culture

    (c) bacterial cell culture

    (d) fungal cell culture

Ans.    (a)

Sol.    Contact inhibition is a special properly of cells in the nature during the cell arrest growth phase when two or more cells come into contact with each other.

    This phenomenon is observed in animal cells culture. Cells like cancerous have lost this property which allows them divide in an uncontrolled manner.

13. Choose the correct set of match between Group I and Group II.

    Group I    Group II

    P.    Igm 1. Present in various body secretions

    Q.    IgE 2. Antigen presentation

    R.    IgA 3. Allergic reaction

    S.    MHC 4. Complement activation

            5. Ten heavy chains and ten light chains

    (a) P–4, Q–3, R–1, S–5

    (b) P–5, Q–3, R–1, S–2

    (c) P–5, Q–3, R–4, S–1

    (d) P–3, Q–2, R–1, S–5

Ans.    (b)

Sol.    The correct match are as follows :

    (i)    IgM is the largest immunoglobulin in the human circulatory system. IgM has a polymeric nature, due to which it possesses high avidity and is particularly effective at complement activation.

    (ii)    IgE is a type of antibody that is present in minute amounts in the body but plays a major role in allergic diseases.

    (iii)    IgA is an antibody that plays critical role in mucosal immunity. In its secretory form IgA is the main immunoglobulin found in mucous secretions, including tears, saliva, sweat, colostrum and secretions from the gastrointestinal tract, genitourinary tract, prostate and respiratory epithelium.

    (iv)    Antigen presentation is a process of the immune system. As immune cells cannot see inside other cells which may be infected with viruses of bacteria. thus, they rely on information conveyed by fragments of intracellular components being presented on MHC molecules on the cell surface.

14. Choose the correct set of match between Group I and Group II

    Group I    Group II

    P.    Gibberellins 1. Breaking dormancy

    Q.    Ethylene 2. Apical dominance

    R.    Cytokines 3. Fruit ripening

    S.    Abscisic acid 4. Seed germination

            5. Cell division and growth

    (a) P–1, Q–3, R–2, S–4

    (b) P–1, Q–2, R–4, S–5

    (c) P–2, Q–3, R–4, S–1

    (d) P–4, Q–3, R–5, S–1

Ans.    (d)

Sol.    The correct match are as follows :

    (i)    Gibberellins of GAs functions to initiate mobilisation of storage materials in seeds during germination.

    (ii)    Ethylene stimulates the growth of fruits in some plants. It is considered responsible for the changes that occur during the ripening of fruits.

    (iii)    Cytokines are a group of chemicals that influence cell division and shoot formation.

    (iv)    Abscisic acid is one of the most important plant growth regulator, as it acts as an inhibitory chemical compound that affects bud growth, seed and bud dormancy.

15. Choose the correct set of match between Group I and Group II

    Group I

    P.    Pyridoxal 5'-phosphate

    Q.    Biotin

    R.    Thiamine pyrophosphate

    S.    N5, N10-methylene tetrahydrofolate

    Group II

1. Carboxylation reaction

2. One-carbon transfer reaction

3. Decarboxylatio reaction

4. Oxidation reduction reaction

    5. C-C bone cleavage

    (a) P–1, Q–3, R–5, S–2

    (b) P–3, Q–1, R–5, S–4

    (c) P–3, Q–1, R–5, S–2

    (d) P–1, Q–4, R–2, S–3

Ans.    (c)

Sol.    The correct match are as follows :

    (i)    Pyridoxal 5-phosphate is the active form of vitamin-B6 which acts as coenzyme in all transmination reactions and in certain decarboxylation, deamination and racemisation reactions of amino acids.

    (ii)    Biotin also known as vitamin H is a coenzyme for carboxylase enzyme, involved in the synthesis of fatty acids, isoleucine and valine and in gluconeogenesis.

    (iii)    Thiamine pyrophosphate is vitamin B1 derivative which catalyses the reversible decarboxylation reaction, i.e. cleavage of a substrate compound at a carbon-carbon bond connecting a carbonyl group to an adjacent reactive group, usually a carboxylic acid or an alcohol reactive group, usually a carboxylic acid or an alcohol in reactions including pyruvate dehydrogenase, alpha ketoglutarate dehydrogenase and transketolase.

    (iv)    N5, N10-methylene tetrahydrofolate is used as an coenzyme in the biosynthesis of thymidine. It is the carbon-1 donor in the reaction catalysed by thymidylase synthase and thymidylate synthase (FAD).

16. Choose the correct set of match between Group I and Group Ii.

    Group I    Group II

    P.    Calcitonin 1. Blood glucose level regulation

    Q.    Glucagon 2. Female reproductive system maintenance

    R.    Adrenalin 3. Mammary gland development

    S.    Prolactin 4. Increase in basal metabolic rate

            5. Calcium homeostasis

    (a) P–5, Q–1, R–4, S–3

    (b) P–2, Q–5, R–3, S–1

    (c) P–5, Q–1, R–2, S–4

    (d) P–4, Q–2, R–3, S–5

Ans.    (a)

Sol.    The correct match are as follows :

    (i)    Calcitonin slows down the activity of osteoclasts found in bone. The decreases blood calcium levels. Thereby helping calcium homeostasis.

    (ii)    Glycoagon is released by pancrease when blood sugar levels fall too low. It causes the liver to convert stored glycogen into glucose, which is released into the bloodstream.

    (iii)    Adrenalin is a hormone produced by the adrenal glands during high stress or exciting situations. Its function is to increase the basal metabolic rate.

    (iv)    Prolactin stimulates the mammary glands to produce milk (Lactation). Increased serum concentration of prolactin durnig pregnancy cause enlargement of the mammary glands of the breasts and prepare for the production of milk.

17. Microaerophilic bacteria have the ability to grow in

    (a) high concentration of oxygen

    (b) absence of oxygen

    (c) low concentration of oxygen

    (d) low concentration of carbondioxide

Ans.    (c)

Sol.    A microaerophile is a microorganisms, can be a bacteria that require oxygen to survive, but require environment containing lower levels of oxygen that are present in the atmosphere (i.e. < 21% O2; typically 2-10% O2).

18. Cocci arranged in the form of chains are classified as

    (a) Sterptococci

    (b) Micrococci

    (c) Sarcinae

    (d) Staphylococci

Ans.    (a)

Sol.    Streptococcus is a genus of coccus gram-positive bacteria. In this bacteria cellular division occurs along a single axis and thus, they grow in chains or pairs. Hence they are given name streptos meaning easily bent or twisted like a chain (twisted chain).

19. The process of Tyndallization requires

    P.    Temperature of 100°C

    Q.    Pressure of 15 psi

    R.    Time period of 30 min

    S.    Free flow of steam

    (a) P, Q, R

    (b) P, Q, S

    (c) P, R, S

    (d) Q, R, S

Ans.    (c)

Sol.    Tyndallisation is heat sterilisation by steaming the food or medium for a few minutes at atmospheric pressure on three or four successive occassions, separated by 12-18 hours intervals of incubation at a temperature favourable for bacterial growth. It was invented by John Tyndall and process includes stema heating ot temperature of 100°C for 30 min.

20. Choose the correct set of match between Group I and Group II

    Group I

    P.    Campylobactor jejuni

    Q.    Neisseria gonorrhoeae

    R.    Pneumocystis carnii

    S.    Haemophilus aegyptius

    Group II

1. eye infection in human

2. lung disease in human with AIDS

3. intenstinal disease with diarrhea

4. sexually transmitted disease

    5. skin infection

    (a) P–3, Q–4, R–2, S–1

    (b) P–3, Q–5, R–2, S–1

    (c) P–2, Q–3, R–4, S–1

    (d) P–2, Q–4, R–1, S–5

Ans.    (a)

Sol.    The correct match are as follows :

    (i)    Campylobactor jejuni is the species of gram-positive bacteria whic his one of the most common cause of human gastroenteritis in the world.

    (ii)    Neisseria gonorrhoeae is a gram-negative bacteria responsible for the sexually transmitted infection gonorrhea.

    (iii)    Pneumocystis camii pneumonai (PCP) is an opportunistic infection that occurs in immunosuppressed populations, i.e. often affects people with HIV and AIDS.

    (iv)    Haemophilus aegyptius is a causative agent of acute and often purulent conjunctivitis, more commonly known as pinke eye.

21. Asexual reproductive process of budding occurs in

    (a) all fungi

    (b) yeasts

    (c) fungi undergoing sexual reproduction

    (d) Bacilus subtilis

Ans.    (b)

Sol.    Yeast is a fungi which reproduce by asexual or sexual reproductive cycles. The most common mode of vegetative growth in yeast is asexual reproduction by budding.

    Where as Bacillus subtilis is a bacterium which divides asexually by binary fission.

22. One mL of E.coli culture was diluted to 100 mL and 0.5 mL of the diluted culture was plated on to agar plate. After 12 h of incubation, 200 colonies were observed. What was the number of bacteria per mL in the original culture?

    (a) 2 × 104

    (b) 4 × 104

    (c) 1 × 105

    (d) 2 × 105

Ans.    (a)

Sol.    For determining the number of bacteria we count the colonies on the surface of tha agar plate and multiply it by the dilution factor. We have 200 colonies on a 1:100 dilution. Thus, 100 × 200 = 2 × 104 colony forming units or 2 × 104 cells or bacteria per mL.

23. Phylogeny describes species

    (a) morphological similarities with other species

    (b) reproductive compatibilities with other species

    (c) evolutionary history

    (d) geographic distribution

Ans.    (c)

Sol.    Phylogenetics is the study of phylogenesis, i.e. the evolutionary history, development and relationships among groups of organies, e.g. species or populations. It implies that different species arise from previous forms via descent and that all organisms from the smallest microbe to the largest plants and vertebrates are connected by the passage of genes along the branches of the phylogenetic tree that links all of life.

24. The term prophage refers to

    (a) an auxotrophic mutant

    (b) a phage DNA incorporated in to bacterial chromosome

    (c) host DNA packed into viral particles

    (d) DNA of lytic phage

Ans.    (b)

Sol.    Pro phase is a stable inherited form of bacteriophage, in which the genetic material of the virus is integrated into, replicated and expressed with the genetic material of the bacterial host.

25. According to Darwin, two different areas within the same continent have different species because they have different

    (a) evolutionary mechanisms

    (b) ancestors

    (c) environments

    (d) evolutionary times

Ans.    (c)

Sol.    According to Darwin, two different areas within the same continent can have species if they are living in different physical environment. Because different environmental conditions imposes different strategies for the species for its survival allowing it to change in its appearance to cope up with the changing environment.

26. A sequence of species through which an organic molecule passes in a community is reffered to as

    (a) pyramid of energy

    (b) food chain

    (c) food web

    (d) nutrient cycle

Ans.    (b)

Sol.    A food chain is the flow of energy from one organism to the next organism, i.e. from lower trophic level to higher trophic level. Thus, a sequence of species through which an organic molecule passes in a community can be referred to as food chain.

27. When a number of genes are transcribed as one mRNA, such mRNA is termed as

    (a) multimeric

    (b) polymeric

    (c) polycistronic

    (d) polysomal

Ans.    (c)

Sol.    An mRNA can be monocistronic or polycistronic. But most of the mRNA found in bacteria and archea are polycistric. In polycistronic mRNA, the mRNA molecule carries the information of several proteins, which are translated into several proteins. They have got a related function and also are grouped and regulated together in an operon.

28. The presence and location of a specific gene in a bacterial genome can be detected by

    (a) Southern blot

    (b) Western blot

    (c) Eastern blot

    (d) Northern blot

Ans.    (a)

Sol.    The presence and location of specific gene in a bacterial genome can be detected by southern blot. As it is a method used in molecular biology for detection of a specific DNA sequence in DNA samples.

    It combines transfer of electrophoresis separated DNA fragments to a fitter membrane and subsequent fragment detection by probe hybridisation.

29. Match the terms in Group I with their definitions in Group II

    Group I

    P.    Ammonification

    Q.    Denitrification

    R.    Nitrification 3

    S.    Nitrogen fixation

    Group II

1. Conversion of atmospheric nitrogen into ammonia

2. Conversion of organic nitrogen into ammonia

3. Conversion of nitrite or nitrate into atmospheric nitrogen

4. Conversion of ammonium into nitrite and nitrate

    (a) P–2, Q–3, R–1, S–4

    (b) P–3, Q–2, R–4, S–1

    (c) P–3, Q–2, R–4, S–1

    (d) P–2, Q–3, R–4, S–1

Ans.    (d)

Sol.    The correct matching for the following events taking place in a nitrogen cycle :

    (i)    Bacteria or fungi convert the organic nitrogen into ammonia by a process called ammonification or mineralisation.

    (ii)    Denitrification is the reduction of nitrates back into the largely inert nitrogen gas completing the nitrogen cycle.

    (iii)    Nitrification is the conversion of ammonia into nitrite and nitrate primarily by soil living bacteria and other nitrifying bacteria.

    (iv)    Nitrogen fixation is a process in which nitrogen in the atmosphere is converted into ammonium or nitrogen dioxide.

30. Nucleosome is composed of

    (a) DNA and histone proteins

    (b) DNA, histone and non-histone proteins

    (c) DNA, RNA and histone proteins

    (d) RNA, histone and non-histone proteins

Ans.    (a)

Sol.    The nucleosome is the fundamental subunits of chromatin. Each nucleosome is made of a little less than two turns of DNA wrapped around a set of eight proteins called histones, which are known as a histone octamer.

31. Usually there is one specific tRNA for each amino acid but some of the amino acids are recognized by more one tRNA. The tRNAs that recognize the same amino acid are known as

    (a) Cognate tRNAs

    (b) Isoaccepting tRNAs

    (c) Isoschizomers

    (d) Catenated tRNAs

Ans.    (b)

Sol.    Isoaccepting tRNAs are group of different tRNAs accepting the same amino acid. The identity of a specific isoaccepting tRNA is determined by particular nucleotide sites within the group. The anticodon may serve as an identity site, however the six serine tRNAs do not share a common anticodon base.

32. Shine-Delgarno sequence is a part of

    (a) Eukaryotic mRNA

    (b) Prokaryotic mRNA

    (c) Eukaryotic tRNA

    (d) Eukaryotic rRNA

Ans.    (b)

Sol.    The shine Dalgarno (SD) sequence is a ribosomal binding site in prokaryotic messenger RNA generally located around 8 bases upstream of the start codon AUG. The RNA sequence helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning the ribosome with the start codon.

33. Which of the following statements are FALSE about Palindromes?

    P.    DNA which reads the same sequence from both directions but in antiparallel orientation

    Q.    DNA which reads the same sequence from both directions but in parallel orientation

    R.    It is recognized by a specific restriction endonuclease and causes specific cleavage

    S.    It is recognized by exonucleases and causes non-specific cleavage.

    (a) P and R

    (b) P and S

    (c) Q and R

    (d) Q and S

Ans.    (a)

Sol.    Palindromic sequences are those nucleotide sequences which are the same when read in either direction, usually consisting of adjacent inverted repeats. Restriction endonuclease recogenition sites are palindromes.

34. Which of the following statements are true regarding DNA replication?

    P.    It is semiconservative both in prokaryotes and eukaryotes

    Q.    It is semiconservative in eukaryotes but conservative in prokaryotes

    R.    Both leading and lagging strands are replicated by DNA Pol-III in prokaryotes

    S.    Leading and lagging strands are replicated by two different polymerases in eukaryotes

    (a) P, R, S

    (b) P, Q, R

    (c) Q, R, S

    (d) P, Q, S

Ans.    (a)

Sol.    DNA replication is the process of producing two identical replicas from one original DNA molecule. DNA is made up of two strands and each strand of the original DNA molecule serve as a template for the production of the complementary strand, a process known as semiconservative replication which is in both eukaryotes and prokaryotes.

    In prokaryotes both leading and lagging strands are replicated by DNA poly-III. But in eukaryotes leading strand is synthesised by DNA polymerase and and lagging strand by DNA polymerase .

35. Which one of the following statements are true about genetic code and translation?

    P.    Genetic code is degenerate because more than one codon codes for a particular amino acid

    Q.    Genetic code is degenerate because a single codon codes for more than one amino acid

    R.    Genetic code degeneracy is due to wobble nature of 3' base

    S.    Fidelity exists in translation as there is no proof reading mecahnism

    (a) P, Q, S

    (b) Q, R, S

    (c) P, Q, R

    (d) P, R, S

Ans.    (a)

Sol.    Transiation is the synthesis of proteins directed by a mRNA template.

    The genetic code is degenerative, i.e. several codons code for the same amino acid and this degeneracy is due to wobble nature of 3' base. This has an effect on the number of different tRNA molecules needed for protein synthesis. Fidelity exists in mRNA translation as proof reading occurs during protein synthesis.

36. Which of the following techniques are used for transfer of a gene into the cells?

    P.    Electroporation

    Q.    Electoelution

    R.    Particle bombardment

    S.    Microinjection

    (a) Q, R, S

    (b) P, Q, R

    (c) P, R, S

    (d) P, Q, S

Ans.    (c)

Sol.    Except electroelution all other methods like electroporation, particles bombardment and microinjection are used for transfer of a gene into the cells. This is because electroelution is a method used to extract a nucleic acid or a protein sample from an electrophoresis gel. We use this method to recover DNA fragments from a particular region of agrose or polyacrylamide gel.

37. Match the terms in Group-I with terms in Group-II

    Group-I    Group-II

    P.    RNA-P 1. Iac operon

    Q.    Leucine zipper 2. rRNA gene transcription

    R.    RNA Pol-I 3. tRNA gene transcription

    S.    Attenuation 4. Tarnscription factors

            5. Ribozymes

            6. trp operon

            7. mRNA splicing

    (a) P–7, Q–5, R–3, S–1

    (b) P–4, Q–5, R–2, S–1

    (c) P–5, Q–4, R–2, S–6

    (d) P–4, Q–5, R–3, S–6

Ans.    (c)

Sol.    The correct match are as follows :

    (i)    RNA-P is DNA dependent RNA polymerase, i.e. an enzyme that produces primary transcript RNA. It is a ribosomal enzyme or a ribozyme.

    (ii)    Leucine zippers are a dimerisation domain of the b ZIP (Basic region leucine Zipper) class of eukaryotic transcription factors.

    (iii)    RNA polymerase I is in higher eukaryotes and is a polymerase that only transcribes ribosomal RNA.

    (iv)    Attenuation is a mechanism of control in some bacterial operons which was discovered by Charles Yanofsky in the trp operon in E.coli.

38. Which one of the following modifications leads to protein degradation?

    (a) Methylation

    (b) Acetylation

    (c) Phosphorylation

    (d) Ubhiquitination

Ans.    (d)

Sol.    Ubiquitination refers to the post translational modification of a protein by the covalent attachment by isopeptide bond with one or more ubiquitin monomers. The most unique function of ubiquitin is to label proteins for proteasomal degradation.

39. Which one of the following protein is involved in the nucleation step of microtubules in vivo?





Ans.    (d)

Sol.    -Tubulin is an important part of multiprotein complex that nucleates the minus end of microtubules. It is found primarily in centrosomes and spindle pole bodies, since these are the areas of most abundant microtubule nucleation.

40. If a codon in mRNA is UAC, the anticodon on tRNA will be

    (a) 5'AUG3'

    (b) 5'GUA3'

    (c) 5'ATC3'

    (d) 5'CTA3'

Ans.    (b)

Sol.    If a codon in mRNA is UAC, the anticodon on tRNA will be 5' GUA 3'. As adenine makes double bond with uracil and guanine makes triple bonds with cytosine.

41. Which one of the following structure-function pairs is NOT correct?    

    (a) Nucleolus-rRNA sysnthesis

    (b) Lysosome-intracellualr digestion

    (c) Endoplasmic reticulum-glycosylation

    (d) Microtubules-muscle contraction

Ans.    (d)

Sol.    The incorrect pair is of microtubules. Because microtubules are filamentous intracellular structures that are responsible for various kinds of movements in all eukaryotic cells. They are involved in nucleic and cell division, organisation of intracellular structure and intracellular transport as well as ciliary and flagellar motility.

42. Lysosomal protein targeting takes place through

    (a) COP-coated vesicles

    (b) Clathrin coated vesicles

    (c) Liposome

    (d) Receptor mediated endocytosis

Ans.    (b)

Sol.    Lysosomal enzymes are targeted into the lysosome by formation of clathrin coated vesicles at the trans Golgi network.

43. The release of Ca2+ from endoplasmic reticulum to cytoplasm in response to stimulus is mediated by

    (a) cAMP

    (b) IP3

    (c) DAG

    (d) Calmodulin

Ans.    (b)

Sol.    The endoplasmic reticulum (ER) is the major site where protein folding occurs, it is the site of stress resulting from unfolded proteins. Under too much stress, the cell opens calcium channels on the endoplasmic reticulum by making IP3 molecules. These molecules bind to the IP3 receptors on the ER and cause a release of calcium ion into the cytosol. The calcium helps turn on a group of proteins that digest the cell from the inside.

44. Match the terms in Group-I with terms in Group-II

    Group-I    Group-II

    P.    Leucoplast 1. Protein modification and targeting

    Q.    Mitochondria 2. Microtubule organizing centre

    R.    Golgi complex 3. Starch storage

    S.    Centriole 4. Kreb's cycle

            5. Glycogen storage

            6. Calvin cycle

    (a) P–3, Q–4, R–2, S–1

    (b) P–5, Q–4, R–6, S–5

    (c) P–3, Q–6, R–4, S–5

    (d) P–3, Q–4, R–1, S–2

Ans.    (d)

Sol.    The correct match for the following are :

    (i)    Leucoplast are colourless plastids which function as storage places for starch.

    (ii)    In eukaryotic cells, the kreb's cycle or citric acid cycle occurs in the matrix of the mitochondrion.

    (iii)    The Golgi complex function as a factory in which proteins received from the ER are further processed and sorted for transport to their eventual destinations, i.e. lysosomes, the plasma membrane or secretion.

    (iv)    The centroles together with the pericentriolar matrix around it are called the microtubule organising centre (MTOC).

45. At constant pressure, the internal energy of a gaseous system will always decrease for

    (a) an endothermic process with decrease in the volume

    (b) an endothermic process with increase in the volume

    (c) an exothermic process with decrease in the volume

    (d) an exothermic process with increase in the volume

Ans.    (d)

Sol.    We know,




46. First ionization energy of C, N, O and Si follows the order

    (a) Si < O < N < C

    (b) C < N < O < Si

    (c) Si < C < N < O

    (d) Si < C < O < N

Ans.    (d)

Sol.    First ionisation energy is defined as the energy required to excite an electron from outermost shell.


    Hence, the correct order is N > O > C > Si

47. Which one of the following isoelectronic ions has the largest ionic radius?

    (a) O2–

    (b) F

    (c) Mg2+

    (d) Na+

Ans.    (a)

Sol.    For isoelectronic ions, radii of cation < radii of anion. Due to decrease in effective nuclear charge of anion, size of anion increases.

        ionic radii follow the trend as Mg2+ < Na+ < F < C2–

48. The correct set of match between molecules of Group I and their shapes in Group II is

    Group I    Group II

    P.    I3 1. Square pyramidal

    Q.    H2S 2. Trigonal bipyramidal

    R.    XeOF4 3. Linear

    S.    PCl5 4. Angular

    (a) P–3, Q–4, R–1, S–2

    (b) P–4, Q–3, R–1, S–2

    (c) P–3, Q–4, R–2, S–1

    (d) P–4, Q–3, R–2, S–1

Ans.    (a)



49. Thallium (Tl) exhibits monovalency whereas aluminium (Al) exhibits trivalency. This is due to

    (a) the energy required to unpair outer s-electrons in Tl exceeds the energy involved in the bond formation

    (b) Tl has only one electron in its outermost orbital

    (c) Al can use its vacant d-orbitals for the bond formation

    (d) Tl is a non-metal

Ans.    (a)

Sol.    Thallium exhibits monovalency whereas aluminium (Al) exhibits trivalency due to the energy required to remove and S-electron in Tl exceeds the energy involved in the bond formation.

50. Which one of the following compounds has non-zero spin only magnetic moment?

    (a) [Fe(CN)6]4–

    (b) [Co(NH3)6]3+

    (c) [Zn(H2O)6]2+

    (d) [NiF6]4+

Ans.    (d)

Sol.    For non-zero spin only magnetic moment there must be presence of unpaired electrons.

    In [Fe(CN)6]4– = Fe2+ = 3d6; due to CN



51. Isomerism exhibited by the pari of compounds [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] is

    (a) linkage

    (b) coordination

    (c) ionization

    (d) geometric

Ans.    (b)

Sol.    The pair of compounds present below form coordination isomerism, in which two complexes interchange their oxidation states as well coordination spheres.

    [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]

52. With increase in pressure, the equilibrium concentration of product will NOT change for





Ans.    (c)

Sol.    With increase in pressure, the equilibrium concentration of product will not change in those reactions, where number of moles of gaseous reactant is equal to number of moles of gaseous product.




53. During a cyclic process, which one of the following is NOT always zero?

    (a) Enthalpy change

    (b) Entropy change

    (c) Internal energy change

    (d) Work done by the system

Ans.    (d)

Sol.    During a cyclic process, path function does not always give zero. Since, work done is a path function,


54. The rate of reaction (r) is expressed as, rr = k[A]m[B]n. The rate constant (k) for this reaction is 2L2 mol–2s–1. The possible values of m and n are

    (a) 1 and 1

    (b) 1 and 2

    (c) 1 and 3

    (d) 1 and 4

Ans.    (b)

Sol.    r = k[A]m[B]n

    Order = m + n

    Unit of rate constant (k ) = L2 mol–2 s–1

    Comparing it from generalised formula for nth order.

        k = (mol)1 – n Ln – 1 s–1

    n = 3

        k = (mol)–2 L2 s–1

    m + n = 3

        m = 1, n = 2

55. Half-cell reaction for the electrode Ag/AgCl/Cl is





Ans.    (c)

Sol.    Electronde : Ag|AgCl| Cl


56. Which one of the following is NOT a correct statement for carbohydrates?

    (a) Epimers give the same osazone

    (b) D(+)-glucose undergoes mutarotation

    (c) -D(+)-glucose and -D(+)-glucose are anomers

    (d) Conversion of -D(+)-glucose to -D(+)-glucose is called suugar inversion

Ans.    (d)

Sol.    Glucose and mannose are epimers to each other which form same osazone.



    (b)    D(+)-glucose undergoes mutarotation.

    (c)    -D(+)-glucose and -D(+)-glucose are anomers, because they differ in configuration at C1.

    (d)    Conversion of (+)-sucrose to D(+)-glucose and D(–)-fructose is known as sugar inversion.

57. In an electophoresis experiment at pH 5 (shown below) x, y and z refer respectively to


    (a) Lysine, alanine and asparitc acid

    (b) Alanine, aspartic acid and lysine

    (c) Lysine, aspartic acid and alanine

    (d) Aspartic acid, alanine and lysine

Ans.    (a)

Sol.    In electrophoresis experiment at pH = 5.

    X = Lysine

    Y = Alanine

    Z = Aspartic acid

58. Which one of the following 0.1 M solutions has the lowest pH?

    (a) NaNO2

    (b) NH4Cl

    (c) NaCl

    (d) NH3

Ans.    (b)

Sol.    Lowest pH value = More acidic

        Strong acid + Weak base

        HCl = NH4OH

        NH4Cl + H2O

59. In a thin layer chromatrogarphy experiment, three spots k, l and m are detected in an iodine chamber. The sports k, l and m respectively are


    (a) PhCH2OH, PhCOOH, PhCh2OCOCH3

    (b) PhCOOH, PhCH2OCOH3, PhCH2OH

    (c) PhCOOH, PhCH2OH, PhCH2OCOCH3

    (d) PhCH = OCOCH3, PhCH2OH, PhCOOH

Ans.    (a)

Sol.    When the compuonds that differ in polarity are present in the mixture, the more polar compound has a stronger interaction with the silica and is more capable to dispel the mobile phase from the binding places. As a consequences, the less polar compound moves higher up the plate resulting in the highest position.

    So, the order is


60. IR strentching frequency at ~2200, ~1700, ~1100 and ~1600 cm–1 corresponds respectively to the functional groups





Ans.    (a)

Sol.    IR stretching frequency for


61. Which one of the following species does NOT have 6π electrons?





Ans.    (b)




62. Which one of the following compounds will NOT show three signals in its 1H NMR spectrum?

    (a) CH3CH2CH2Br    


    (c) CH3OCH2CH2OCH3    


Ans.    (c)



    In above compounds, there are equivalent protons are present which resonate differently in same 1HNMR spectroscopy and gives three different signals.


    In this compound, there are two equivalent protons which give two signals.

63. Choose the correct match between reactions of Group-i and named reactions in Gorup-II.






1. Clemmensen reduction

2. Chotten-Bauuman reaction

3. Aldol condensation

4. Hoffman degradation

    5. Reimer-Tieman reaction

    6. Sandmever reaction

    (a) P–3, Q–2, R–1, S–5

    (b) P–2, Q–3, R–4, S–5

    (c) P–3, Q–4, R–5, S–6

    (d) P–4, Q–3, R–6, S–5

Ans.    (a)



64. (+)-Mandelic acid has a specific rotation of +160°. What is the observed specific rotation of a mixture of 40%(–) mandelic acid and 60% (+)-mandelic acid?

    (a) –32%

    (b) +32°

    (c) +64°

    (d) –64°

Ans.    (b)

Sol.    Observed specific rotation


65. Group-I lists fundamental forces in nature and Group-II lists the particles relevant to these forces. Chooose the correct set of match.

    Group-I    Group-II

    P.    Gravitational 1. Charges

    Q.    Electromangetic 2. Nucleons

    R.    Weak nuclear 3. Masses

    S.    Strong nuclear 4. Elemeentary particles

    (a) P–1, Q–4, R–2, S–3

    (b) P–3, Q–1, R–2, S–4

    (c) P–4, Q–2, R–3, S–1

    (d) P–3, Q–1, R–4, S–2

Ans.    (d)

Sol.    P.    Gravitational force depends upon masses.


    Q.    Electromagnetic force is created due to electric and magnetic fields (i.e. charges).

    R.    Weak nuclear force is related with elementary particles.

    S.    Strong nuclear force is due to changes in the nucleus i.e. nucleons.

66. A small body of mass 0.2 kg unergoes a uniform circular motino on a frictionless horizontal surface. The body is attached to the centre by a string of length 2m and has a linear speed of 10 m/s. The force exerted by the string on the mass is

    (a) 1N

    (b) 5N

    (c) 10N

    (d) 50N

Ans.    (c)

Sol.    Force exerted by the string on the mass will be equal to the tension (T) on the string.


67. Resistance (R), capacitance (C) and inductance (L) are connected in series in a circuit. Keeping the resonant frequency same, the quality factor (Q) can be doubled if

    (a) L is increased to 2L and C is decreased to C/2

    (b) C is increased to 2C and L is decreased to L/2

    (c) L is increased to 4L and C is decreased to C/4

    (d) C is increased to 4C and L is decreased to L/4

Ans.    (a)

Sol.    Quality factor of L-C-R series circuit is given by




68. de-Boglie wavelengths of two electrons which start from rest and accelerated by potentials V and 4V are and respectively. The ratio is

    (a) 1 : 2

    (b) 1 : 4

    (c) 2 : 1

    (d) 4 : 1

Ans.    (c)

Sol.    Kinetic energies of electrons can be calculated as


    where, potentials are V1 = V and V2.v1 and v2 are speeds of electrons.

    de-Brogle wavelengths can be written as



69. The mass numbers of two nuclei M and N are 4 and 8 respectively. The ratio of the volumes of the nuclei, MN : VV is

    (a) 1 : 2

    (b) 1 : 4

    (c) 1 : 8

    (d) 1 : 16

Ans.    (a)

Sol.    The relation between radius of nucleus and mass number is given by

        R = R0(A)1/3

    where, R0 is a constant and A is mass number of the nucleus.

    Volume of the nucleus is



70. A student is interested in converting a galvanometer into a voltmeter. The student should

    (a) connect a large resistance in series with the galvanometer

    (b) connect a large resistance in parallel to the galvanometer

    (c) connect a small resistance in series with the galvanometer

    (d) connect a small resistance in parallel to the galvanometer

Ans.    (a)

Sol.    To convert a galvanometer into a voltmeter a large resistance (R) must be connected in series with the galvanometer (G).


71. The phase difference between points that are 2m apart along the direction of propagation of a wavelength of 6m is

    (a) 60°

    (b) 120°

    (c) 150°

    (d) 180°

Ans.    (b)

Sol.    The relation between phase difference (ΔΦ) and path difference is given by



72. A car moving at a constant speed of 36 km/h in the direction of wind and assited by the flow of wind which imparts a force of 50N. The frictional force between the types and the road is 100N. The engine power required is

    (a) 50W

    (b) 100W

    (c) 500W

    (d) 1800W

Ans.    (c)

Sol.    Given that,

    Speed of the car, v = 36 km/h = 36 × = 10 m/s

    Force imparted by the wind, tW = 50 N

    Frictional force, f = 100 N

    Net force on the car = f - fW

        Fnet = 100 – 50 = 50 N

    This force is resistive in nature.

    The engine power required to overcome this resistive force and move the car with the constant velocity is

        P = (Fnet) (V) = (50) (10) = 500 W

73. The centres of two planets (P and Q) are at a distance L apart and the ratio of their masses is 1 : 4. What is the distance between the centre of the lighter planet (P) and the point on the line PQ at which the net gravitational force is zero?


    (a) L/4

    (b) L/3

    (c) L/2

    (d) 3L/4

Ans.    (b)

Sol.    Consider the planets A and B having masses in the ratio mA : mB = 1 : 4 as shown in the figure :


    Suppose that at point P net gravitational force is zero.




74. A charge is place on a solid conductor. Under static condition, which one of the following statements is FALSE?

    (a) There is no free charge in the interior of the conductor

    (b) Potential is constant over the surface of the conductor

    (c) Electric field is zero inside the conductor

    (d) Electric field at the surface has both norma and tangential components

Ans.    (d)

Sol.    Electric field due to the charge on the solid conductor is perpendicular to the sphere as shown below. So, option (d) is incorrect.


75. A circular wire of radius R1 carrying a current I in the anticlockwise directions is concentric with another circular wire of radius R2(R2 < R1) also carrying a current I in the clockwise direction as shown in the figure

    The magnetic field at the centre is






Ans.    (a)

Sol.    Magnetic field due to a circular current carrying conductor at the centre is given by [symbols have their usual meaning] So, net magnetic field due to the current carrying circular conductor is

    B = B1 – B2 = (According thumb rate direction of magnetic field will perpendicular to plane upward)

76. The ionization energy for a hydrogen atom in its first excited state (n = 2) is

    (a) 13.6 eV

    (b) 3.4 eV

    (c) –3.4 eV

    (d) –13.6 eV

Ans.    (c)

Sol.    Energy of the hydrogen atom in nth state is given by


77. The volume expanxsion coefficient for a uniform solid cube is and the linear expansion coefficinet is . For small temperature changes, the relationship between and is





Ans.    (c)

Sol.    Given,

        Volume expansion coefficient =

        Linear expansion coefficient =

        For small temperature changes.

    Change in length a cube can be written as


    Change in volume of the cube




78. The velocity (v) of a particle moving along positive x-axis is given by v = where k is a positive constant. At time 0 = t the particle is at x = 0. The distance of the particle as a function of time is given by

    (a) x = kt1/2

    (b) x = k2t

    (c) x = kt3/2

    (d) x = k2t2

Ans.    (d)

Sol.    Given that,





79. Two springs of spring constants, k1 and k2 are connected in series where one end is fixed to a wall and on the end is connected to a block of mass m. The arrangement is kept on a frictional surface. What is the frequency of oscillation when the mass is sightly displaced?






Ans.    (a)

Sol.    Equivalent spring constant of the series combination of the spring is given by keq =


    Time period of oscillation of the block



80. L, M and N are points on the isotherms (T1 and T2) as shown in the figure. If WLM, WMN and WLN denote the work done by one mole of an ideal gas along the paths LM, MN and LN respectively, then


    (a) WLN > WLM > WMN

    (b) WLM > WMN > WLN

    (c) WLN > WMN > WLM

    (d) WMN > WLN > WLM

Ans.    (c)

Sol.    Work done during the process LN is

        WLN = area (LNPQ)


    Similarly, work done during the process MN

        WMN = area (MNPQ)

    As    area (LNPQ) > area (MNPQ)

        WLN > WMN

        LM is an isochoric process

        WLM = 0

        WLM > WMN > WLM

81. In Si single crystal, the intrinsic carrier concentration ni at temperature T1 is twice that of at temperature T2. Then for an n-type Si crystal,

    (a) electron concentration n at T1 is twice to that of n at T2

    (b) electron concentration n at T1 is four times to that of n at T2

    (c) hole concentration n at T1 is twice to that of n at T2

    (d) hole concentration n at T1 is four times to that of n at T2

Ans.    (b)

Sol.    For an n-type Si crystal, majority carrier is electron.

    As we know that,


    For temperature T1


    For temperature T2

    From Eqs. (i) and (ii)


    where, n1 and n2 are electron concentrations at temperature T1 and T2.


82. The minimum value of as shown in the figure for which total internal reflection occurs at the interface between liquid and air is (refractive indices of the media are given in brackets in the figure and n1 > n2 > n3)






Ans.    (b)

Sol.    For total internal reflection at the surface CD, the angle of incidence at CD surface should be critical angle

    Applying Snell's law,




83. If the vectros coplanar, then the value of p is

    (a) –2

    (b) –1

    (c) 1

    (d) 2

Ans.    (a)


    For coplanarity [a b c] = 0

         = 0

        1(–p + 2) – 1(2p + 2) – 1(4 + 2) = 0

        – p + 2 – 2p – 2 – 6 = 0

        – 3p – 6 = 0

        – 3p = 6

        p = – 2

    Hence, the required value of p be –2.

84. A committee of 4 members is to be formed out of 6 men and 4 women. If the committee has to include at least 2 women and a particular woman is always selected, the number of ways it can be formed is

    (a) 36

    (b) 60

    (c) 64

    (d) 90

Ans.    (c)

Sol.    P (at least 2 women and a woman is always selected)

        = P (2W) + P (3W) + P (4W)

        = 3C1.6C2 + 3C2.6C1 + 3C3

        = 3 × + 3 × 6 + 1

        = 45 + 18 + 1 = 64

85. The shortes distance of the point (1, 0, 1) from the straight line given by is

    (a) 2

    (b) 2

    (c) 3

    (d) 3

Ans.    (c)

Sol.    We know that shortest distance of the point (x1, y1, z1) from the straight line given by = is

        d =

    Now, shortest distance of the point (1, 0, 1) from the straight line



86. The area of the region in the first quadrant bonded by the curves 2yx2 and 3yx2 is

    (a) 1/12

    (b) 1/6

    (c) 1/2

    (d) 3/4

Ans.    (a)

Sol.    Area of the region in first quadrant bounded by y = x2 and y = x3




87. The value of is

    (a) –1

    (b) e

    (c) 1

    (d) 0

Ans.    (d)





    (a) 0

    (b) –1

    (c) 1/14

    (d) 1

Ans.    (b)




89. If tan A and tan B are the roots of the equation x2 + px + q = 0 then the value of tan(A + B) is

    (a) p/q

    (b) q/p

    (c) q–(1–p)

    (d) p/(1–q)

Ans.    (d)

Sol.    Given, tan A and tan B are the roots of equation

    x2 – px + q = 0




90. A fair coin is tossed 100 times. The probabilit of getting tails an odd number of times is

    (a) 1/8

    (b) 1/4

    (c) 3/8

    (d) 1/2

Ans.    (d)

Sol.    P (getting tails an odd number of times)




91. If y is a function of x given at (0, 0) is

    (a) –1

    (b) 0

    (c) 1


Ans.    (a)






92. If a is given by a = , then the value of a e–a is

    (a) 1/2

    (b) 2/3

    (c) 1

    (d) 3/2

Ans.    (b)





93. The complex number equals

    (a) –1

    (b) 1



Ans.    (a)





    (a) –1

    (b) 0

    (c) 1

    (d) 2

Ans.    (b)






95. The distance of the point (1/2, 0) and the line of intersection of the circles x2 + y2 = 4 and (x + 1)2 + y2 = 4 is



    (c) 1

    (d) 2

Ans.    (c)

Sol.    Equation of the line of intersection of two given circles

    x2 + y2 – 4 = 0 and (x + 1)2 + y2 – 4 = 0 is

    [(x + 1)2 + y2 – 4] – [x2 + y2 – 4] = 0

        (x + 1)2 + y2 – 4 – x2 – y2 + 4 = 0

        x2 + 1 + 2x – x2 = 0

        2x + 1 = 0

    Now distance of from line 2x + 1 = 0 is of



96. A missile is projected from the ground at an angle of 45° with the vertical. If it has to hit a target 50m high at a horizontal distance of 100 m, the velocity of projection is (g is the acceleration due to gravity)

    (a) a2

    (b) 2a2

    (c) 3a2

    (d) 4a2

Ans.    (b)

Sol.    Given, a be the radius of circle inscribed a rectangle








    Hence, the maximum area of rectangle inscribed in a circle is 2a2.

97. A missile is projected from the ground at an angle of 45° with the vertical. If it has to hit a target 50m high at a horizontal distance of 100m, the velocity of projection is (g is the acceleration due to gravity)





Ans.    (d)

Sol.    Equation of path of projectile is








98. The maximum value of 1 2 3 5 x X subject to the constraints

    0 <x1< 4

    0 <x2< 6

    3x1 + 2x2 < 18 is

    (a) 21

    (b) 27

    (c) 30

    (d) 36

Ans.    (d)

Sol.    Given, 0 < 4, 0 < x2< 6 and 3x1 + 2x2< 18

    Let 3x1 + 2x2 = 18



    Let    Z = 3x1 + 5x2

    At (4, 0)

        Z = 3 × 4 + 5 × 0 = 12

    At (4, 3)

        Z = 3 × 4 + 5 × 3 = 12 + 15 = 27

    At (2, 6)

        Z = 3 × 2 + 5 × 6 = 6 + 30 = 36

    Hence, the maximum value of 3x1 + 5x2 is 36.

99. The function y(x) satisfies the differential equation is

    (a) In2


    (c) 1

    (d) e

Ans.    (b)


    On integrating both sides w.r.t 'x', we have










100. The sum of the series


    (a) 2

    (b) 0.5

    (c) 1

    (d) 0.25s

Ans.    (c)

Sol.    Given,



    = 1