21.    Which one of the following graphs best describes the dependence of free energy change of ATP hydrolysis on Mg²⁺ concentration ?

    (a)

    (b)

    (c)

    (d)

Ans.    (a)

Sol.    

    

    Hence, correct option is (A).

22.    Which one of the following statements is true regardeing amino acids ?

    (a)    Proline has high propensity to form -helix in globular proteins.

    (b)    Both isoleucine and threonine can exist as diastereomers.

    (c)    Side chain pK_a of aspartic acid is more than the side chain pK_a of glutamic acid.

    (d)    The dihedral angle of proline is more restricted than the dihedral angle.

Ans.    (b)

Sol.    The stereochemistry of the most of the standard amino acids is defined by the two possible mirror image isomers or enantiomers. Two of the standard amino acids, isoleucine and threonine have a second asymmetric carbon in addition to their -carbons namely their -carbons. The stereochemistry of these two amino acids is thus defined by 4 stereo isomers i.e., two enantiomers as well as two diastereoisomers (non-mirror image stereoisomers) as in the popup windows for isoleucine and threonine.

23.    The interaction energy between two opposite charges separated by 3Å in vacuum is kJmol^–1. The interaction energy between these two charges in water will be closest to

    (a)    –1500 kJmol^–1

    (b)    –166 kJmol^–1

    (c)    –55 kJmol^–1

    (d)    –6 kJmol^–1

Ans.    (d)

Sol.    

    

    Hence, the interaction between two opposite charges in water is closed to –6 kJ mol^–1.

24.    Match the following vitamins with the corresponding pathological conditions aising from their deficiencies.

    P.    A    1.    Pernicious anaemia

    Q.    B₁₂    2.    Subdermal haemorrhaging

    R.    D    3.    Night blindness

    S.    K    4.    Rickets

    (a)    P–3, Q–1, R–4, S–2

    (b)    P–3, Q–2, R–4, S–1

    (c)    P–3, Q–1, R–2, S–4

    (d)    P–4, Q–1, R–2, S–4

Ans.    (a)

Sol.    Vitamin A : Vitamin A is an essential vitamin needed for growth and development, cell recognition vision, immune function and reproduction.

    Vitamin A deficiency can result from inadequate intake, fat malabsorption or liver disorders. Deficiency impairs immunity and hematopoiesis and causes rashes and typical ocular effects e.g., xeropthalmia and night blindness.

    Vitamin B₁₂ : Pernicious aneuria, which makes it hard for your body to absorb vitamin B₁₂.

    Vitamin D deficiency in children can cause rickets. It has links to high BP in children.

    Vitamin K : It plays a key role in helping blood clotting, preventing excessive bleeding. Low levels of vitamin K can raise the risk of uncontrolled bleeding. While vitamin K deficiencies are in adults, they are very common in new born infants. Early hemorrhagic disease of the new born is a disease resulting from vitamin K deficiency.

    Hence, A. (a)-3, (b)-1, (c)-4, (d)-2.

25.    In a form of stress response, bacteria synthesize a group of proteins called stress proteins (or heat shock proteins) such as DnaK, DnaJ, GroEL, GroES, and GrpE. DnaK is an ATP binding protein, which attahces to the newly in the subsequent process of protein folding?

    (a)    The affinity of DnaK to the polypeptide increases upon hydrolysis of the ATP to ADP.

    (b)    DnaJ is an exchange factor that replaces ADP with ATK in a DnaK.

    (c)    ATP hydrolysis is required for the phosphorylation of DnaJ.

    (d)    ATP hydrolysis is required for the phosphorylation of GrpE.

Ans.    (a)

Sol.    The polypeptide binding and release cycle of the molecular chaperone DnaK (Hsp70) of E. coli is regulated by the two-chaperons DnaJ and GrpE.

    Here, the DnaJ conversion of DnaK. ATP (T state) to DnaK. Pi (R state) as monitored by intrinsic protein fluorescence, is monophasic and occurs simultaneously with ATP hydrolysis.

    This ins in contrast with the T R conversion in absence of DnaJ which is biphasic, the first phase occuring simultaneously with the hydrolysis of ATP.

    Therefore, the affinity of DnaK to the polypeptide increaes upon hydrolysis of the ATP to ADP.

26.    Given beow is the [P] vs time plot of an enzymatic reaction carried out by the enzyme 'x'. Which one of the following statements is the correct interpretation of the data ?

    (a)    The K_m and V_max of the enzyme 'x' are 15 and 60 units, respectively.

    (b)    The V_max is 60 but the K_m cannot be determined.

    (c)    The K_m is 15 but the V_max cannot be determined.

    (d)    Neither the K_m nor the V_max of the enzyme 'x' can be determined from the these data.

Ans.    (d)

Sol.    Km and Vmax are determined by incubating the enzyme with varying concentrations of substrate; the results can be plotted as a graph of rate of reaction (v) against concentration of substrate ([S], and will normally yield a hyperbolic curve.

27.    A form and Z form of double stranded DNA differ in the handedness of their helics, nucleotide sequences, and configuration of base to sugar. Based on these properties, which one of the following statements defines a correct combination for A and Z form of DNA?

    (a)    Right handed double helix and anti-configuration for the base of the sugar arrangement in A DNA and left handed double helix with alternating sequence of G and C (as a general pattern), and alternating syn and anti-configuration for the base to sugar arrangement in the Z DNA.

    (b)    Right handed double helix and syn-configuration ofr the base to sugar arrangement in A DNA; and left handed double helix with alternating A and G sequece (as a general pattern) and anti-configurations for base to sugar arrangement in the Z DNA.

    (c)    Left handed double helix and anti-configuration for base to sugar arrangement in the A form DNA and right handed double helix and syn-configuration for base to sugar arrangement in the Z form DNA.

    (d)    Left handed double helix and syn-configuration for base to sugar arrangement in the A form DNA and right handed double helix and anti-configuration for the base to sugar arrangement for the Z form DNA.

Ans.    (a)

Sol.    A DNA is a right-handed double helix fairly similar to the more common B-DNA form, but with a shorter, more compact helical structure whose base pairs are not perpendicular to the helix-axis as in B-DNA. Z-DNA is one of the many possible double helical structures of DNA. It is a left-handed double helical structure in which the helix winds to the left in a zigzag pattern, instead of to the right, like the more common B-DNA form.

28.    The following statements are made :

    P.    B form of DNA has ~10 base pairs/turn and A form of DNA has ~2.3Å helix rise per base pair.

    Q.    Both the A and B form of DNA have wider major groove and narrow groove.

    R.    The crystalline nature of cellulose is brought about by (1 4) linkage between the glucose subunits.

    S.    The double bonds in natural lipids are always cis, which provides fluidity to the plasma membrane.

    Which of the following combiantions represent the correct statements ?

    (a)    P and R

    (b)    Q and R

    (c)    P and S

    (d)    R and S

Ans.    (c)

Sol.    One helical turn of B-DNA contains about 10.5 base pairs that are buried inside the helix and are almost perpendicular to the helical axis. Naturally occurring fatty acids are generally in the cis configuration, which provide fluidity to plasma membrane.

29.    The following observations are made on a 30-residues polypeptide :

    P.    Unordered structuer is observed in water but a helical conformation is observed in medium of low dielectric constant.

    Q.    The peptide is resistance to degradation by proteases.

    R.    Red blood cells are lysed by the peptide.

    S.    -mercaptoethanol has no effect on peptide structure.

    Which of the following statements can be correctly attributed to the above obserrvations?

    (a)    The peptide is entirely composed of D-amino acids and is amphipathic.

    (b)    The peptide is entirely composed of L-amino acids and is not amphipathic.

    (c)    The peptide is rich in disulphide bonds between D-cysteines.

    (d)    The peptide is entierly composed of L-armoatic amino acids.

Ans.    (a)

Sol.    The peptide is entirely composed of D- amino acids and is amphipathic.

30.    Which one of the following is not a G-proteins coupled receptor?

    (a)    Epinephrine receptor

    (b)    Transferrin receptor

    (c)    Glucagon receptor

    (d)    Thyroid stimulating hormone receptor

Ans.    (b)

Sol.    Epinephrine receptor : It is a sympamomimetic catecholamine that exerts into pharmacologic effects on both and -adrenergic receptors using G-protein-linked second messenger system.

    Transferring receptor : It is a carrier protein for transferrin. It is needed for the import of iron into the cell and is regulated in response to intracellular iron concentration.

    Glucagon receptor : The glucagon receptor is a G-protein coupled, seven transmembrane domain receptor and is expressed in the liver, pancreatic isletc, heart, adipose tissue and less abundantly in other tissues.

    Thyroid stimulating hormone receptor : It is a receptor that responds to thyroid sitmulating hormone and stimulates the production of thyroxine and tridothyronine. The TSH receptor is a member of the G-protein coupled receptor superfamily of integral membrane proteins and is coupled to the G protein.

    Hence, transferrine receptor is not a G-protein coupled receptor.

    Thus, correct answer is (b).

31.    In a neuron, proteins and membranes aer prirmarily synthesized in the cell body. The materials must be transported down the axon to the synaptic region using microtubules in an anterograde fashion. Such axonal transport is directed by

    (a)    Dynein

    (b)    Kinesin I

    (c)    Dynein and Kinesin I

    (d)    Myosin

Ans.    (b)

Sol.    Kinesin motor proteins move cargos along microtubules away from the cell body (anterograde transport), whereas dynein together with the dynactin complex move cargos in the retrograde direction toward the cell body.

32.    For bacterial growth, a single cell elongates in size before it divides into two, in a process called binary fission. During cell growth,

    (a)    new peptidoglycan synthesis is required along with the hydrolysis of bonds linking the old peptidoglycan chains.

    (b)    new polypeptideglycan synthesis is required but no hydrolysis of the old peptidoglycan occurs.

    (c)    the old peptidoglycan is completely degraded and replaced with the newly synthesized longer polymer.

    (d)    newly synthesized peptidoglycan is utilized to deposit a new layer of the peptidoglycan in the cell wall.

Ans.    (a)

Sol.    A single cell elongates in size before it devides into two cells, is called binary fission, for bacterial growth. New peptidoglycan sythesis is required alongwith the hydrolysis of bonds linking the old peptidoglycan chains during the cell growth.

33.    Phosphoenol pyruvate : sugar phosphotransferase system (PTS) transports a variety of sugars into bacteria. In E. coli, PTS consists of EI, EII (EIIA, EIIB and EIIC) and Hpr. During this process the sugar molecule is phosphorylated by direct transfer of phosphate goup from

    (a)    EI-P

    (b)    EIIA-P

    (c)    EIIB-P

    (d)    Hpr-P

Ans.    (c)

Sol.    The bacterial phosphoenolypyrvate (PEP) : Sugar phosphotransferse system (PTS) mediates the uptake and phosphorylation of carbohydrates and is involved in signal transduction. It comprises two general phosphotransferase proteins (EI and HPr) and a species dependent variable number of sugar specific depenedent variable number of sugars specific enzyme II complexes (IIA, IIB, IIC). E1 and HPr transfer phosphoryl groups from PEP to the IIA units. IIA and IIB frequently transfer phosphates to the sugar, which is translocated by the IIC unit. During this process, the sugar molecule is phosphorylated by direct transfer of phosphate group from EIIB-P.

    

34.    Some statements regarding the process of autophagy are given below :

    P.    Autophagy occurs when cells contain aggregated proteins.

    Q.    Autophagosomes fuse with any organelles.

    R.    Autophagosome is a singel membrane structure.

    S.    Autophagosomes fuse with lysosomes to fomr autophagolysomes.

    Which one of the following combination of the above statements is correct ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    S and P

Ans.    (d)

Sol.    During selective autophagy, the autophagosomal membrane forms around a specific cargo that may contain aggregated cytosolic proteins, virtually any types of intracellular organelles marked for degradation, or invading pathogens such as bacteria and viruses. In the autophagosome-lysosome fusion step, the outer lipid bilayer membrane of autophagosomes fuses with the lipid bilayer membrane of lysosomes. Since SNARE proteins are essential for the fusion of lipid bilayers, these proteins are required for autophagosome-lysosome fusion.

35.    The nucleosome is the fundamental subnit of chromatin in eukaryotes. Following statements are made about nucleosome :

    1.    Generally, a typical nucleosome contains ~200 bp of DNA and two copies of each histone (H2A, H2B, H3 and H4).

    2.    146 bp length of DNA per nucleosome core particle is strictly maintained across the organisms.

    3.    The histone octamers are not conserved during/after replication, howeve, H3₂–H4₂ tetamers are.

    4.    Variants have been identified for all core histones except histon H3.

    5.    While wrapping around the core histones, the structure of DNA is altered at the middle of the nucleosome core particle and exhibits increased number of base pairs per tun.

    Which one of the following combination of statements is most appropriate ?

    (a)    1, 3 and 5

    (b)    1, 2 and 4

    (c)    2, 4 and 5

    (d)    1, 3 and 4

Ans.    (a)

Sol.    A single nucleosome consists of about 150 base pairs of DNA sequence wrapped around a core of histone proteins. wrapped around a core of histone proteins and two copies of each histone. This is accomplished by wrapping the DNA around structural histone proteins, which act as scaffolding for the DNA to be coiled around.

36.    The membrane phospholipid structures in bacteia and archea differ. Which one of the following correctly states the differences between the two ?

    (a)    The bacterial membrane phospholipid consist of D-glycerol linked to hydrophobic chains (tails) with ester bonds whereas those of the archeal membrane consist of L-glycerol linked to hydrophobic tails through ether bonds.

    (b)    The bacterial membrane phospholipid consist of L-glycerol linked to hydrophobic chains (tails) with ester bonds whereas those of the archeal membrane consist of D-glycerol linked to hydrophobic tails through ether bonds.

    (c)    The bacterial membrane phospholipid consist of D-glycerol linked to hydrophobic chains (tails) with ester bonds whereas those of the archeal membrane consist of L-glycerol linked to hydrophobic tails through ester bonds.

    (d)    The bacterial membrane phospholipid consist of L-glycerol linked to hydrophobic chains (tails) with ester bonds whereas those of the archeal membrane consist of D-glycerol linked to hydrophobic tails through ester bonds.

Ans.    (a)

Sol.    Bacteria and archaea differ in the lipid composition of their cell membrane and the characteristics of the cell wall. Bacterial cell walls contain peptidoglycan, but archean cell walls do not have peptidoglycan, they have polysaccharides, glycoproteins or protein-based cell walls. The membrane phospholipid structures in bacteria and archea differ, because the bacterial membrane phospholipid consist of D-glycerol linked to hydrophobic chains with ester bonds whereas those of the archeal membranes consists of L-glycerol linked to hydrophobic tails through ether bonds.

37.    Thousands of proteisn taht are synthesized in the cytoplasm are imported into the nucleus through the nuclear pore complex (NPC) every minute. These proteisn contain Nuclear Localization Signal (NLS) that direct their selective transport into the nucleus. This nuclear import requires :

    P.    A small momeric G-protein Ran.

    Q.    A nuclear transport receptor that interacts with the NLS on a cargo protein.

    R.    A GTPase activating potein (GAP) bound to the chromatin tethered to the nuclear membrane.

    S.    A Guanine Exchange Fator (GEF) bound to the chromatin inside the nucleus.

    Which one of the following options represents all correct statements ?

    (a)    P, R, S

    (b)    P, Q, S

    (c)    Q, R only

    (d)    P, R only

Ans.    (b)

Sol.    Transport through nuclear pores

    

    Thousands of proteins are synthesized in the cytoplams and are imported into the nucleus through NPC every minute. These proteins contain NLS that direct their selective transport in the nucleus, the nuclear import requires a small monomeric G protein ram, nuclear transport receptor that interacts with the NLS on a cargo protein and guanine exchange factor (GEF) bond to chromatin inside the nucleus.

38.    Individual in a population are divided into various groups (in column 'X') based on the set of enzymes they have (in column 'Y') :

    Column X        Column Y

        A    1.    Fucose transferase

        B    2.    GalNAc transferase

        AB    3.    Gal transferase

        O

    Which one of the following combinations is not correct for the group type and the set of enzymes a persone has ?

    (a)    A-(1) and (2)

    (b)    B-(1) and (3)

    (c)    AB-(1), (2) and (3)

    (d)    O-(1), (2) and (3)

Ans.    (d)

Sol.    In the formation of A and B antigens, H molecule (coded by H gene) acts as a precursor. H molecule (antigen) is found in every individual. In case of A antigen, H contains an additional N-acetylgalactosamine (GalNAc). Similarly, in case of B antigen an additional galactose (Gal) is linked to H molecule The addition of the terminal GalNAc to the H molecule is catalyzed by GalNAc transferase, a product of A gene. Similarly, the product of the B gene is the Gal transferase adding the Gal residue to the H molecule. Individuals of type AB possess both enzymes and thus have two A and B antigens and individuals of type O contain only H molecule.

    So, (d) is incorrect statement.

39.    Cells are phsyically linked to one another and to extracellular matrix through their cytoskeleton and this imparts strength and rigidity of tissues and organs. Most of the animal cells have three types of cytoskeletal filaments, which are listed in column A. The possible functions are listed in column B.

        Column A                Column B

    P.    Intermediated filaments    1.    Determine the shape of cell surface and are necessary for cell locomotion.

    Q.    Microtubules    2.    Maintain the position of membrane-enclosed organelles and provide intracellular transport

    R.    Actin filaments    3.    Provide mechanical strength of a cell

    Which one is the correct match ?

    (a)    P-1, Q-2, R-3

    (b)    P-2, Q-1, R-3

    (c)    P-3, Q-2, R-1

    (d)    P-3, Q-1, R-2

Ans.    (c)

Sol.    Intermediate filaments provide mechanical support for the plasma membrane where it comes into contact with other cells.

    Microtubules are responsible for the variety of cell movements, including the intracellular transport and positioning of membrane vesicles and organelles. It maintains the position of membrane enclosed organelles and provide intracellular transport.

    Actin filaments are particularly abundent beneath the plasma membrane, where they form a network that provides mechanical support, determines cell shape and allows movement of the cell surface.

    Hence, (a)-3, (b)-2, (c)-1.

40.    During cell cycle, entry in the S-phase is tightly regualted. This is possible because :

    P.    APC/C promotes ubiquitination of S-phase cyclins and mitotic cyclins, marking them for proteolyses at the mitotic exit.

    Q.    Cyclin B1 helps in the activation of S-phase CDKs only ni late G1.

    R.    As mitotic CDK activity declines in late mitosis, cdc14 phosphatase activates APC/C by dephosphorylating Cdh1, thus promoting formation of APC/C^Cdn1.

    S.    Securin keeps S-phase cyclins in inactive state till late G1.

    Which one of the optioons represents all correct statements ?

    (a)    P and Q

    (b)    P and R

    (c)    Q and R

    (d)    Q and S

Ans.    (b)

Sol.    

    Because APC/C promotes ubiquitination of S-phase cyclins and mitotic cyclins, marking them for proteolyses at the mitotic exit and as mitotic CDK activity declines in late mitosis, cdc14 phosphatease activates APC/C by dephosphorylating Cdh1. Hence promoting formation of APC/cdh^–1.

41.    Which one of the following statements related to transcription and processing of mRNA is incorrect ?

    (a)    During prokaryotic transcription, DNA binding properties of RNA polymerase are altered by sigma factor.

    (b)    In eukaryotic transcription, synthesis of rRNA, mRNA and some small RNAs occurs by RNA polymerase I, II and III respectively.

    (c)    Splicing observed in tRNA involves successive/sequential cleavage and ligation reactions while pre-mNA splicing proceeds through lariat formation.

    (d)    mRNAs with premature stop codons are degraded by Nonsense-Mediated Decay (NMD) and mRNAs without an in-frame stop codon get accumulated and translated in the cytoplasm.

Ans.    (d)

Sol.    mRNAs containing premature termination codons (PTCs) are known to be degraded via nonsense mediated mRNA decay (NMD). mRNAs containing any type of PTCs (UAA, UAG and UGA) are detained in the nucleus, whereas their wild type counterparts are rapidly exported. Our data indicates that translating ribosomes in the nucleus the frame and detect the PTCs in the nucleus. Morever, the shuttling NMD protein Upf₁, specifically associates with PTC + mRNAs (PTC – containing mRNAs) in the nuclecus and is required for nuclear retention of PTC + mRNAs, together our data working model that PTCs are recognized in the nucleus by translating ribosomes/resulting in recruitment of Upf₁, which in turn functions in nuclear retention of PTC + mRNA. Nuclear PTC recognition adds a new layer of mRNA and may be vital for ensuring the extraordinary fidelity required for the protein production.

    Hence, option (D) is not correct.

42.    The amino acid side chains of the four histones in the nucleosome are subjected to remarkable variety of post-translation modifications such as phosphorylation, acetylation and methylation. Which one of the following post-translational marks on histone tails is usually associated with transcriptional repression ?

    (a)    Acetylation of H3K9

    (b)    Methylation of H3K9

    (c)    Acetylation of H4K5

    (d)    Phosphorylation of H3S10

Ans.    (b)

Sol.    Acetylation Transcription (ON)

    Phosphorylation ON

    Methylation OFF

    Deacylation OFF

    Repression mean of method 2

    Contradictory element methylation

    Methylation of H3K9 is the post-translational marks on histone tails is usually associated with transcriptional repression.

43.    In context of DNA methylation, which one of the following statements is false ?

    (a)    Generally, methylation occurs at the 3^rd carbon position of cytosine and convets it to 3-methylcytosine.

    (b)    Maintenance methyltransferase acts constitutively of hemimethyalted sites and converts them to fully methylated sites.

    (c)    During mammalian gametogenesis, the genomic methylation patterns are erased in primordial germ cells.

    (d)    Replication converts a fully methylated site to hemimethyalted site.

Ans.    (a)

Sol.    C S – Methyl Cytogene

    

44.    Which one of the following regulatoy proteins can act as a positive and negative regulator on binding to the same DNA elements ?

    (a)    Lac repressor (LacI)

    (b)    Lambda (cI) repressor

    (c)    Ara C protein (AraC)

    (d)    Trp repressor (TrpR)

Ans.    (b)

Sol.    Lambda (Cl) repressor, which negatively regulates two promotes to block Cytic gene expression. The expression of Cl is itself controlled by positive and negative feedback as Cl binds to O_R to regulate the P_RM promotor. In addition, to direct interactions with operator DNA. Cl tetramers band at O_L with operator DNA. Cl tetramers band at O_L and O_R can come together to form an ocfamer, looping the DNA that lies between them and allowing O_L to assist with negative regulation of P_RM.

    We used a fluorescent reporter protein to measure the Cl concentration for a set of constructs that differ in their ability to assume various forms of the looped structure. Based on the observed steady state fluorescence for these constructs, the presence of O_L increases P_RM activation unless both operators can be fully occupied.

    When DNA is looped, P_RM activation can be 2 to 4 fold higher that is possible for unlooped DNA. Therefore, lambda (Cl) repressor is the regulatory proteins can act as a positive and negative regulator on binding to the same DNA elements.

45.    Following statements were made about the post-transcriptional processing of RNA in eukaryotes.

    1.    Soon after transcriptional initiation, RNA polymerase II pauses ~30 nucleotides downstream from the site of initiation until the Cap structure is added to the 5' end of the nascent pre-mRNA.

    2.    The 5' splice sites are functionally divergent whereas the 3' sites are functionally equivalent.

    3.    In addition to helping in recognition of the splice sites, the exon definition also functions as a splicing regulator by allowing pairing and linking of adjacent 5' and 3' splice sites.

    4.    The intron definition mechanism applies only to the large itnrons (about 500 nucleoties length) and assists in achieving alternate splicing.

    5.    The splicing reactions carried out in vitro have revealed that the first and second transesterification reactions are reversible.

    Which one of the following combination of statements is correct ?

    (a)    1, 2 and 4

    (b)    2, 3 and 4

    (c)    2, 4 and 5

    (d)    1, 3 and 5

Ans.    (d)

Sol.    Soon after transcriptional initiation, RNA polymerase II pauses ~30 nucleotides downstream from the site of initiation until the Cap structure is added to the 5' end of the nascent pre-mRNA. In addition to helping in recognition of the splice sites, the exon definition also functions as a splicing regulator by allowing pairing and linking of adjacent 5' and 3' splice sites. The splicing reactions carried out in vitro have revealed that the first and second transesterification reactions are reversible.

46.    The figure below shows the structure of a replication fork.

    Based on the information, following statements are made :

    P.    (i) represents the leading strand while (ii), (iii) and (iv) represent the Okazaki fragments.

    Q.    Among the Okazaki fragments, synthesis of (iv) occurs prior to the synthesis of (iii) and (ii).

    R.    Among the Okazaki fragments, synthesis of (ii) occurs prior to the synthesis of (iii) and (iv).

    Which one of the following options represents the correct statements(s)?

    (a)    P only

    (b)    Q only

    (c)    P and Q

    (d)    P and R

Ans.    (c)

Sol.    

    Therefore (i) represents the leading strand while (ii), (iii) and (iv) represents the Okazaki fragments. Among the okazaki fragments, synthesis of (iv) occurs prior to the synthesis of (iii) and (ii).

    Hence, correct option is (c).

47.    Many organisms encode only 18 aminoacyl-tRNA synthetases (aaRS). These organisms lack aaRS that use Asn or Gln (as one of the substrates) for direct aminoacylation of the tRNA^Asn and tRNA^Gln, respectively. Which one of the following statements represents the correct option ?

    (a)    The organisms lacking AsnRS and GlnS lack Asn and Gln in their proteins.

    (b)    In these organisms, selected Asp and Glu residue in the proteins are post-translationally modified by a regulated mechanism.

    (c)    In these organisms, the tRNA^Asn and tRNA^Gln are first aminoacylated by AspRS and GluRS, respectively, and then the Asp and Glu attahced to the tRNAs are modified to Asn and Gln, respectively.

    (d)    In these organisms, the precursors of mRNAs that encode AspRS and GluRS are alternatively spliced to generated AsnRS and GlnRS.

Ans.    (c)

Sol.    

    Therefore, tRNA^Asn and tRNA^Gln both are the first aminoacylated by AspRS and GluRS.

48.    Following statements are made about double-strand break repair (DSBR) model of homologous recombination :

    1.    Process of DSBR recombination is triggered by introducing a doubl-strand break in a DNA duplex.

    2.    In a process known as 3'-end resection, the exonucleases along with a DNA helicase degrade one strand on either side of the break and generates 3'-single stranded termini.

    3.    One strand of the donor duplex is displaced due to formation of heteroduplex DNA and generates a displacemetn loop (D-Loop).

    4.    Branch migration allows the point of crossover to move in 5' 3' direction of recipient strand.

    5.    Completion of DSBR recombination may geneates either crossover recombinant or non-recombinant product.

    Which one of the following combination of statements is correct ?

    (a)    1, 2 and 4

    (b)    1, 3 and 5

    (c)    1, 3 and 4

    (d)    1, 2 and 5

Ans.    (b)

Sol.    Process of DSBR recombination is triggered by introducing a doubl-strand break in a DNA duplex. One strand of the donor duplex is displaced due to formation of heteroduplex DNA and generates a displacemetn loop (D-Loop). Completion of DSBR recombination may geneates either crossover recombinant or non-recombinant product.

49.    In the diagram below, the dotted line marks the point of initiation of bidirectional replication.

    P.    On the right side of the dotted line, leading strand synthesis occurs using the upper strand as the template.

    Q.    On the right side of dotted line, leading strand synthesis occus using the lower strand as the template.

    R.    A ligase deficient (lig^–) mutant would effect replication of the upper strand on the left side of the dotted line.

    S.    A ligase deficient (lig^–) mutant would affect replication of the lower strand on the left side of the dotted line.

    Which one of the following options represents the combinations of the correct statements?

    (a)    P and S

    (b)    Q and R

    (c)    Q and S

    (d)    P and R

Ans.    (c)

Sol.    By observing the diagram given in question, it is observed that On the right side of dotted line, leading strand synthesis occus using the lower strand as the template and A ligase deficient (lig^–) mutant would affect replication of the lower strand on the left side of the dotted line.

50.    Deletion analysis of a promoter region of a gene was carried out to identify the regulatory elements in it. In the figure below, the filled boxes denote the areas of deletion and the observed activities (in arbitrary units) of the promoter are as shown.

    

    Based on the observations, following statements were made :

    P.    The region between –100 and –50 houses a positive regulatory element.

    Q.    The region between –200 and–250 houses a negative regulatory element.

    R.    The region between –150 and –200 houses a positive regulatory element.

    Which one of the following options represents the correct interpretation of the data ?

    (a)    Both P and Q

    (b)    P only

    (c)    Q only

    (d)    Both Q and R

Ans.    (c)

Sol.    We have,

    250 and –300    No site (positive or negative)

    200 and –250    Activity increases negative regulatory site

    150 and –200    (No change)

    150 and –100    Unique region positive regulatory site

    –200 and –250    Negative regualtory site

    100 – 150    Positive regulatory site

    0 – 50    core

    Thus, the region between –200 and –250 houses a negative regulator element.

51.    Following statements were made with respect to transcription in eukaryotes :

    1.    RNA polymerase III synthessi mRNAs in the nucleoplasm.

    2.    The target promoter for RNA polymerase III is usually repesented by a biparite sequence downstream of the transcription start site.

    3.    The assembly factors TFIIIA and TFIIIC assist the bindind of the positioning factor TFIIIB at the precise location.

    4.    TFIIIB is the last factor that joins the initiation complex.

    5.    Phosphorylated Ser residues in the C-terminal domain (CTD) of RNA polymerase II serve as binding sites for mRNA processign enzymes.

    Which one of the following options represents the correct combinaton of the statements ?

    (a)    1, 2 and 3

    (b)    2, 3 and 5

    (c)    2, 4 and 5

    (d)    1, 3 and 5

Ans.    (b)

Sol.    RNA polymerse III

    Type I promoter (Found 5 SrRNA gene)

    

    The target promoter for RNA polymerase III is usually represented by a pibartite seqeunce downstream of the transcription start site. The assembly factors TF IIIA and TF III C assist that binding of the positioning facstor. TF III B at the precise location and phosphorylated ser residues in the C-terminal domain (CTD) of RNA polymerase II serve as binding sites for mRNA processing enzymes were made with respect to transcription in eukaryotes.

52.    Ribosomes prepared from a bacterium were fractionated by sucrose density gradient centrifugation (panel i) to separate the 30S, 50S and 70S populations. When the ribosome preparation was incubated individually with either elongation fator-G (EF-G) or a newly identified protein X, or GTP, the profile remained unchaged. Likewise, no changes were seen in the profile when the ribosomal preparation was incubated with EF-G + GTP. However, when the ribosomal preparation was incubated with proteins X, EF-G and GTP together, it resulted in a change of profile which showed a decrease of the 70S peak area and increase in the peak areas for 30S and 50S (panel ii)

    

    Choose the option that defines a correct conclusion from the observations.

    (a)    Protein X is an anti-association factor which functions in the presence of EF-G and GTP.

    (b)    Protein X is a dissocation factor which functions in the presence of EF-G and GTP.

    (c)    Protein X binds GTP.

    (d)    EF-G is known to bind GTP, hence it can be concluded that the effect of GTP is through EF-G and protein X does not bind GTP.

Ans.    (b)

Sol.    By observing the graph and information mentioned in question, Protein X is a dissocation factor which functions in the presence of EF-G and GTP.

53.    Which one of the following lipid soluble hormones can interact with a cell surface receptor ?

    (a)    Pogesterone

    (b)    Estradiol

    (c)    Thyroxine

    (d)    Prostaglandin

Ans.    (d)

Sol.    The primary lipid soluble hormones that bind to cell surface receptors are the prostaglandins. In both vertebrates and invertebrates, prostaglandins are synthesized and secreted continuously by many types of cells and rapidly broken down by enzymes in body fluids.

    Therefore, prostaglandin hormone is the lipid-soluble hormone, which can be interacted with a cell surface receptor.

54.    Two strains of mice which are gentically identical except for a single genetic locus or region are said to be

    (a)    syngenic

    (b)    allogenic

    (c)    congenic

    (d)    heterogenic

Ans.    (c)

Sol.    A congenic strain is an inbred strain that contains a small genetic region ideally a single gene from another strain but which is otherwise identical to the original inbred strain. Therefore, congenics are the two strains of mice that are genetically identical except for a single locus or region.

55.    Which one of the following synaptic vesicle proteisn is involved in tethering of the vesicle to the cytoskeletal system in the nerve terminus ?

    (a)    Synaptotagmin

    (b)    Synapsin

    (c)    Synaptophysin

    (d)    Synaptobrevin

Ans.    (b)

Sol.    Synapsins are the family of abundant evolutionary conserved phosphoproteins found at nearly all synapses in the nervous system. At presynaptic terminals, they reversibly associate with the surface of synaptic vesicles and presumbly with cytoskeletal proteins such as actin.

    These interaction of synapsins are regualted by phosphorylation at multiple sites. So, synapsins are the synaptic vesicles proteins which are involved in tethering of the vesicle to cytoskeletal system in the nerve terminus.

56.    In what respect does the genome of slow-acting retroviruses differ from those of transducing viruses ?

    (a)    They cannot activate nearby cellular proto-oncogenes after integration into the genome of the host cell.

    (b)    They lack of oncogene.

    (c)    They exclude mouse mammary tumor viruses.

    (d)    They have acquired mutations during acquisition of an oncogene.

Ans.    (b)

Sol.    In the case of retroviruses, the transduction of oncogenes refers to the ability of the retroviruses to acquire cellular proto-oncogenes as integral parts of their genomes and then express those transduced genes as viral oncogenes in another cell within the context of viral replication. Slow-acting transforming retroviruses are as a rule replication-competent and do not transduce oncogenes. Instead, they cause transformation by integrating within or in the vicinity of cellular genes and altering their expression.

57.    Bacterial infectiosn are generally divided into two broad classes : intracellular and etracellular bacterail infectiosn. Given below are some of the properties which are applicable for bacterial infections.

    P.    Humoal immune response is the main protective response against extracellula bacteria.

    Q.    Innate immunity is not effective against intracellular bacterial pathogens.

    R.    Bacterial endotoins do not induce an innate immune response.

    S.    Intracellular bacterial infections generally induce a cell-mediated immune response resulting in secretion of cytokines which activate macrophages.

    Which one of the following combination of statements is correct ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    P and S

Ans.    (d)

Sol.    The extracellular spaces are protected by the humoral immune response, in which antibodies produced by B cells cause the destruction of extracellular microorganisms and prevent the spread of intracellular infections. The activation of B cells and their differentiation into antibody-secreting plasma cells. A number of studies have reported the ability of cytokines to bind to bacteria, which can, in some cases, affect the growth of the organisms. This binding may occur when the organism is either outside or inside a host cell. Macrophages act as scavengers that engulf and destroy microbes, particulate matters or altered host cells, while alerting the immune system through the secretion of cytokines, chemokines and lipid mediators. In addition, macrophages contribute to wound healing and tissue repair processes.

58.    Listeria is a food-borne pathogen that causes mild gasto-intestinal symptoms. To move from one host-cell to another, it polymerizes actin into a comet tail like structue. Listeria can assemble host-cell actin at its rear and because :

    P.    Listeria has on its surface a protein called Act A.

    Q.    Listeria can activate Arp 2/3 complex.

    R.    Listeria has on its surface -tubulin.

    S.    Listeria has on its surfae myosine II motor.

    Which one of the following options represents all correct statements ?

    (a)    P and S

    (b)    P and Q

    (c)    P and R

    (d)    R and S

Ans.    (b)

Sol.    Listeria is a good borne pathogen that causes mild gastro-intestinal symptoms. It can assemble host cell action at its rear end because listeria has on its surface a protein called Act A and it can activate Arp 2/3 complex.

    

59.    Acetylcholine is a potent neurotansmitter, which is released fom the neurons. After release they diffuse across the synaptic cleft and combine with nictonic acetylcholine recepto molecules in the membranes of the postsynaptic cell. The interaction of acetylchlonie with the nicotinic acetylcholine receptor produces large transient incease in the cell pemeability of the membane to specialized ion resulting in signal tansduction for neve impulse. Acetylcholine receptor is a

    (a)    ligand-gated cation channel

    (b)    ligand-gated anion channel

    (c)    voltage-gated cation channel

    (d)    voltage-gated anion channel

Ans.    (a)

Sol.    

    Acetylcholine is a potent neurotranmitter which is released from the neuron. The interaction of acetylcholine with the nictotinic acetylcholine receptor produces large transient increase in the permeability of the membrane. Acetylcholine receptor is a ligand-gated cation channel.

60.    In the nervous system, the action potential is generated at the axon hillock in physiological conditions and it is conducted to the terminal end of axon. The location of specific origin of action potential and its direction-specific conduction are explained by a researcher in the following proposed statements :

    P.    The membrane of axon hillock has highest threshold for the geneation of action potential.

    Q.    The membrane of axon hillock contains large numbers of voltage-gated Na⁺ channels and that makes it more excitable.

    R.    The propagating action potential in the middle of the axon cannot generate another action potential in the direction of cell body since a lage fraction of voltage-gated Na⁺ channels in the perceding portion is voltage inactivated.

    S.    As the number of voltage-gated Na⁺ channels is less in the preceding portion of axonal membrane, the propagating action potential in the middle of the axon cannot generate another action potential in the direction of cell body.

    Which one of the following combinations represents both correct explanations ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    P and S

Ans.    (b)

Sol.    The location specific origin of action potential and its direction specific conduction are explained by researcher in the membrane of axon hillock contains large numbers of voltage gated Na⁺ channels and that makes it more excitable and the propagating action potential in the middle of the axon cannot generate another action potential in the direction of body cell since large fraction in the preceding portion in voltage inactivated.

61.    Strain A mice were crossed with strain B mice and first generation F1 mice were obtained, i.e. (A × B)F1. A scientist then implanted thymectomized and irradiated (A × B)F1 mice with a B-type thymus and then reconstituted the animal's immune system with an intravenous infusion (A × B)F1 bone marrow cells. The chimeric mice were to kill LCMV-infected target cells from the strain A or strain B mice.

    Which one of the following is the correct outcome of the experiment ?

    (a)    LCMV-infected target cells from strain A only will be killed.

    (b)    LCMV-infected target cells from strain B only will be killed.

    (c)    LCMV infected target cells from both strain A and B will be killed.

    (d)    Neither cells from strain A nor from strain B will be killed.

Ans.    (b)

Sol.    LCMV - infected target cells from strain B only will be killed.

62.    Which one of the following plants has a bisporic, 8-nucleate, bipolar embryo sac development ?

    (a)    Oenothera

    (b)    Penaea

    (c)    Plumbago

    (d)    Allium

Ans.    (d)

Sol.    In bisporic embryo sacs, meiosis produces only two megaspores, each containing two haploid nuclei, due to the absence of cytokinesis and cell plate formation following the second meiotic division. The megaspore nearest the micropyle, then undergoes programmed cell death, leaving a single functional megaspore with two haploid nuclei. It is called as Allium type embryo sac.

    Therefore, Allium plant has a bisporic, 8 nucleate bipolar embryo sac development.

    

63.    Centrolecithal eggs show

    (a)    superficial cleavage

    (b)    displaced radial cleavage

    (c)    bilateral cleavage

    (d)    discoidal cleavage

Ans.    (a)

Sol.    In centrolecithal eggs cleavage is meroblastic and superficial, while in telolecithal eggs (birds and fish) cleavage in discoidal.

    Slight angles to one another, so that none of the four cells in one plane of the eight-cell is directly over a cell in the plane.

64.    The following diagram represents a longitudinal section through an Arabidopsis shoot apical meristem (SAM) and leaf primordium at its flank. The dorsal (D) and ventral (V) domains are marked.

    Consider the following statements describing the phenotypes of leaf polarity.

    P.    Loss of D function makes the leaf ventralised whereas its overexpression dorsalizes the leaf.

    Q.    Loss of V function makes the leaf dosalize whereas its overexpression ventralizes the leaf.

    R.    Loss of microRNA miR166 dorsalizes the leaf whereas its overexpression ventralizes the leaf.

    S.    miR166 functions by inhibiting its target mRNA.

    Which one of the following functional models best describes the above results ?

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    The phenotypes of leaf polarity is the loss of D function makes the leaf ventralized where as its overexpression dorsalizes the leaf, loss of V function makes the leaf dorsalized while the overexpression ventralizes the leaf, loss of miRNA, miR166 dorsalize of the leaf ventralizes the leaf and miR166 functions mRNA. Hence correct option is (C).

65.    A "morphogen" can determine the fate of a cell concentration. Given below are some statements on the experiment performed to study the gradient-dependent effect of morphogen, activin on cell fate by placing activin (4 nm)-secreting beads on unspecified cells from an early Xenopus embryo :

    1.    Beads without activin did not elicit expression of either Xbra or goosecoid genes.

    2.    Cells nearest to the beads getting highest concentration of activin indcued goosecoid gene whose prodcut is a transcription factor, specifies the frog's dorsal-most structures.

    3.    Cells nearest to the beads getting highest concentration of activin indced Xbra gene whose product is a transcription factor, specifies the frog's dorsal-most structures.

    4.    Cells farthest from the beads getting negligible activin Xbra gene and become blood vessels and heart.

    5.    Cells farthest from the beads getting negligible activin, activated neither Xbra nor goosecoid and the 'default' gene expression instructed the cells to become blood vessels and heart.

    Which of the above observations and conclusions drawn are correct ?

    (a)    1, 2 and 3

    (b)    2, 3 and 4

    (c)    3, 4 and 5

    (d)    1, 2 and 5

Ans.    (d)

Sol.    Morphogenes act as graded positional cues that control cell fate specification in many developing tissues. This concept, in which a signalling gradient regulates differential gene expression in a concentration dependent manner, provides a basis for understanding many patterning processes.

    

66.    Following statements aer being made regarding specification/determinant during animal development :

    1.    During the course of commitment, the cell may not appear different from its nearest or most distant neighbous in the embryo and show no visible signs of differentiation but its developmental fate is restricted.

    2.    At the stage of specification, cell commitment is not labile.

    3.    A cell or tissue is determined when it is differentiating autonomously even when placed into another region of the embryo, or a cluster of differently specified cells in a petri-dish.

    4.    Cytoskeletal arrangements maintain positioning of nuclei in the syncytium, which enables specification of these nuclei of opposing morphogen gradients namely Bicoid and Caudal in Drosophila.

    5.    Capacity for "mosaic" development allows cells tu acquire different functions as a result of interactions with neighbouring cells.

    Which of the above statements are correct ?

    (a)    1, 2 and 3

    (b)    2, 3 and 4

    (c)    3, 4 and 5

    (d)    1, 3 and 4

Ans.    (d)

Sol.    The commitment of regions of the embryo to form particular tissues or organs is a central concept in development, but the mechanisms controlling this process remain elusive. During commitment, cells may not appear different from its neighbours in embryo and show no visible signs of differentiation but its developmental fate is restricted.

    Autonomous cell fate specification is a form of embryonic specification in which a developing cell is able to differentiate (become a cell carrying out a specialised function) without receiving external signals. Effective syncytium formation for cells with disorganized actin and microtubule cytoskeleton argues against hypothesis that cytoskeleton drives fusion expansion.

67.    During fertilization in mammals proteins Izumo and Juno are required for recognition of sperm and egg. Izumo and Juno are found specifically in sperm and egg, respectively. Which one of the following in vivo experiments will demonstrate that Izumo and Juno interact with each other ?

    (a)    If sperms from a male mouse where Izumo has been knocked out is used to fertilize eggs from a normal female and no fertilization occrus.

    (b)    Whole mount immunostaining for Izumo and Juno shows its presence on the sperm and egg, respectively.

    (c)    If a CFP fused Izumo protein is mixed with YFP fused Juno potein in a tube, FRET occurs, i.e. when CFP is excited, emission of YFP is observed.

    (d)    Two independent kidney cell lines are developed, one expressing Izumo and the other Juno. If the two cells are mixed, they tend to aggregate with each other.

Ans.    (d)

Sol.    Juno–Izumo binding is essential for fertilization and may contribute to the membrane block to polyspermy. Izumo (red) is displayed on the surface of acrosome-reacted sperm and interacts with Juno (green) on the surface of ovulated eggs.

68.    Following statements were made about the events occuring during chick development.

    1.    The fertilized chick egg undergoes discoidal meroblastic cleavage, however the cleavage does not extend into the yolky cytoplasm.

    2.    Development of primary hypoblast is mediated by localized migration of a group of highly specified and connected cluster of 30-40 cells.

    3.    By the stage XIII of the chick embryogenesis and little prior to primitie streak formation, the formation of the hypoblast is just complete.

    4.    Hensen's mode of the chick embryo signifies a region at the anterior end of the primitive streak with regional thickening of cells.

    5.    Inhibition of Wnt planar cell polarity pathway in the epiblast causes the mesoderm and endoderm to form centrally instead of peripherally.

    Which one of the following combinations represents all correct statements ?

    (a)    1, 2 and 4

    (b)    1, 3 and 5

    (c)    1, 2 and 3

    (d)    1, 3 and 4

Ans.    (d)

Sol.    Fertilization of the chick egg occurs in the oviduct, before the albumen and the shell are secreted upon it. The egg is telolecithal (like that of the fish), with a small disc of cytoplasm sitting atop a large yolk. Like fish eggs, the yolky eggs of birds undergo discoidal meroblastic cleavage. In amniotes the mesendoderm cells move inside the embryo through a structure known as the primitive streak, extending from the posterior pole anterior through the midline of the embryo. Primitive streak formation involves large scale cell flows of a layer of highly polarized epithelial epiblast cells. Hensen's node, also called the chordoneural hinge in the tail bud, is a group of cells that constitutes the organizer of the avian embryo and that expresses the gene HNF-3.

69.    The anterior-posterior compartment of each segment of Drosophila is defined by wingless and engrailed genes. The following statements are given towards explaining their regulation :

    1.    Wingless is a secretory factor.

    2.    Engrailed is a secretory factor and forms a long-range concnetration gradient.

    3.    Engrailed regulates Wingless through Hedgehog which forms a short-range concentration gradient.

    4.    -catenin homologue is the signalling molecule upstream of Engrailed, which gets cleaved by GSK3 homologue.

    5.    Cubitus interruptus is an intracellular signalling molecule in the Engrailed expressing cells.

    Which one of the following options has all the correct statements towads the regulation of anterior-posterior compartment of segments ?

    (a)    2 only

    (b)    3 only

    (c)    2 and 5

    (d)    1, 3 and 4

Ans.    (d)

Sol.    The anterior posterior compartment of each segment of Drosophila is defined by wingless and enegrailed genes.

    

    Wingless is a secretory factor, engrailed is a secretory factor and forms a long range concentration gradient and -catenin homologue is the signalling molecule upstream of engrailed.

70.    Vascular wilts are widespread and destructive plant diseases. The symptoms of this disease are primarily caused by the clogging of

    (a)    xylem vessels

    (b)    phloem vessels

    (c)    stomata

    (d)    hydathodes

Ans.    (a)

Sol.    Vascular wilts are caused by bacteria that infect and block xylem vessels of host plants. These deceases are also caused by pathogenic fungi or bacteria that enter the water conducting xylem vessels of a plant, then proliferate within the vessel causing water blockage. This typical symptoms include wilting and death of leaves, followed often by death or serious impairment of whole plant.

71.    Which one of the following reactions takes place during the reduction phase of the Calvin-Benson cycle ?

    (a)    Ribulose 1, 5-biphosphate to 3-phosphoglycerate.

    (b)    1, 3-bisphosphoglycerate to glyceraldehyde-3-sphophate.

    (c)    Dihydroxyacetone phosphate to fructose 1, 6-biphosphate.

    (d)    Ribulose 5-phosphate to ribulose 1, 5-biphosphate.

Ans.    (b)

Sol.    The Calvin cycle reactions can be divided into three main stages: carbon fixation, reduction, and regeneration of the starting molecule.

    Reduction : It is the second stage of Calvin cycle. The 3-PGA molecules created through carbon fixation are converted into molecules of simple sugar – glucose. This stage obtains energy from ATP and NADPH formed during the light-dependent reactions of photosynthesis.

72.    While screening an EMS-mutagenized population of a plant, a researcher identified mutant with redced gibberellic acid sensitivity. Which one of the following proteins is most likely to be defective in this mutant ?

    (a)    Sucose non-fermenting related kinase 2 (SnRK2)

    (b)    Constitutive triple response 1 (CTR1)

    (c)    Phytochrome interacting factor (PIF)

    (d)    Coronative-insensitive 1 (COI1)

Ans.    (c)

Sol.    PIFs predominantly act as constitutive transcriptional activators of genes like PIL1 in the dark (etiolated seedlings), in response to shade ,or at night under diurnal conditions, and light reverses this activity through phy-induced removal or inactivation of the transcription factors. PIFs and GA (a positive regulator of germination) regulate each other's signaling pathways at multiple levels throughout the plant life cycle. PIF1, a strong inhibitor of seed germination in the dark, suppresses GA signaling both directly and indirectly.

73.    Which one of the following statements cannot be included while defining the fermentation process ?

    (a)    Alcohol is formed from sugar residues.

    (b)    Requires an electron transport system.

    (c)    Utilizes an organic compound as the final electron acceptor.

    (d)    Produces lactic acid in oxygen deprived muscle.

Ans.    (b)

Sol.    Fermentation does not involve an electron transport system, and no ATP is made by the fermentation process directly. Fermenters make very little ATP—only two ATP molecules per glucose molecule during glycolysis.

74.    Which one of the following parts of root is involved in perceiving gravity ?

    (a)    Quiescent center

    (b)    Endodermis

    (c)    Root cap

    (d)    Elongation zone

Ans.    (c)

Sol.    The root cap functions in the perception of gravity, the protection of the root apical meristem and facilitation of passage of roots through the soil, but the genes involved in these functions are poorly understood. So, the root cap, a part of root is involved in perceiving gravity.

75.    Which of the following factors is known to be involved in postponing pogrammed cell death in cereal aleurone until endosperm mobilization is complete ?

    (a)    Gibberellic acid

    (b)    Abscisic acid

    (c)    Acidic pH of the vacuoles

    (d)    cGMP mediated signal transduction pathway

Ans.    (b)

Sol.    Abscisic acid (ABA) slows down the process of aleurone cell death and isolated aleurone protoplasts can be kept alive in media containing ABA for up to 6 months. Cell death is bareley aleurone occurs only after cells become highly vacuolated and is mainfested in an abrupt loss of plasma membrnae integrity. Aleurone cell death does not follow the apoptotic pathway found in many animal cells. The hallmarks of apoptosis, including internucleosomal DNA cleavage, plasma membrane and nuclear blebbing and formation of apoptotic bodies are not observed in dying aleurone cells.

    

    So, Abscisic acid is known to be involved in postponing programmed cell death in cereal aleurone endosperm mobilization is completed.

76.    Which one of the following graphs best represents the net CO₂ fixation of typical C₃ and C₄ plants under increasing CO₂ concentration and saturating light ?

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    The chloroplasts of these plants are dimorphic and unlike C3 plants the leaves of C4 plants possess kranz anatomy. It has been suggested that increased [CO2] will increase water use efficiency (WUE) of C3 species, because it causes a reduction in transpiration rate and an increase in CO2 assimilation rate of the plants. In C3 plants, Carbon dioxide fixation is slow and it is faster in C4 plants.

77.    Consider the following facts :

    P.    Chlorophyll absorbs more in the red region of the visible spectrum than in far-red.

    Q.    The phytochrome photoreceptor (P) of the plants occus in two inter-convertible forms, P_r and P_fr where red light converts P_r to P_fr and far-red light converts P_fr to P_r.

    R.    Growing a sun plant under the canopy shed causes increased stem elongation.

    Which one of the following combination of statements is correct for the plants growig under the canopy as compared to those growing above the canopy ?

    (a)    Red : far-red ratio is lower; P_r : P_fr ratio is higher; P_fr inhibits stem elongation.

    (b)    Red : far-red ratio is higher; P_r : P_fr ratio is higher; P_fr inhibits stem elongation.

    (c)    Red : far-red ratio is lower; P_r : P_fr ratio is lower; P_fr promotes stem elongation.

    (d)    Red : far-red ratio is higher; P_r : P_fr ratio is lower; P_fr promotes stem elongation.

Ans.    (a)

Sol.    A low R/Fr ratio indicates high enrichment of the Fr light spectrum in the plant canopy. During the day, the Pr form is converted to Pfr, restoring the Pr/Pfr equilibrium. Therefore, during night periods the Pr/Pfr ratio is high while during daytime the ratio it is low. So plants growing under canopy will have high Pr to Pfr ratio. An inductive effect of Pfr inhibits a late stage in internode expansion, and a phytochrome reaction which operates only in light (and may involve pigment cycling) promotes an early stage of internode development. Stem elongation is thus a function both of the daily duration of light and its red/red+far-red content.

78.    

    Match the above columns involving plant hormones and their signalling pathways :

    (a)    P-1-i, Q-2-ii

    (b)    P-2-ii, Q-1-i

    (c)    P-2-i, Q-1-ii

    (d)    P-1-ii, Q-2-i

Ans.    (b)

Sol.    It has been postulated that plants have two types of auxin and GA receptors, including soluble and membrane-bound forms. AUXIN RESPONSE FACTOR7 integrates gibberellin and auxin signaling via interactions between DELLA and AUX/IAA proteins to regulate cambial activity in plants. Their response is mediated by proteasome- mediated protein degradation. Cytokinin and Brassinosteroid have transmembrane receptor and response is mediated by phosphorylation and dephosphorylation.

79.    The following statements were made with the assumption that the concentration of 3-phosphoglycerate is high inside chloroplasts of an actively photosynthesizing leaf.

    P.    There will be high concentration of triose phosphate in the chloroplast.

    Q.    The activity of ADP-glucose pyrophorylase will be inhibited.

    R.    The carbon flow will be diverted from sucrose to starch.

    S.    Starch synthesis will be inhibited and carbon flow will be more towards sucrose synthesis.

    Which one of the following combinations of above statements is correct ?

    (a)    P and Q

    (b)    Q and S

    (c)    R and S

    (d)    P and R

Ans.    (d)

Sol.    

80.    Certain plant species prodce cyanogenic glycosides to protect them from pathogens. A researchers has identified a variant of such a plant that has higher level of cyanogenic glycoside yet it is highly ssceptible to a specific fungal pathogen. To interpret this counter-intutive observation, the researcher hypothesizes that the fungal pathogen has higher level of

    P.    -glucosidase activity

    Q.    formamide hydrolyase activity

    R.    cytochrome P-450 enzyme

    S.    cyanide-resistant, alternative oxidase activity

    Which one of the following combinations of the above hypotheses is correct ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    Q and S

Ans.    (d)

Sol.    In plants, cyanides are usually bound to sugar molecules in the form of cyanogenic glycosides and defend the plant against herbivores. Phytopathogenic fungi need to secrete different hydrolytic enzymes to break down complex polysaccharides in the plant cell wall in order to enter the host and develop the disease. Fungi produce various types of cell wall degrading enzymes (CWDEs) during infection. Most of the characterized CWDEs belong to glycoside hydrolases (GHs). Alternative respiratory pathway- A pathway for cellular respiration occurring in many plants that permits electron transport and reduction of oxygen to water in the presence of cyanide and other substances that completely inhibit respiration in animals.

81.    The table below photosynthetic type, temperature and sunlight intensity levels.

    

    Which of the following correctly matches the plant photosynthetic type with the temperatue and sunlight conditions in which photosynthetic rate per unit leaf area is maximum for that plant ?

    (a)    P-1-i, Q-3-iii

    (b)    P-3-i, Q-1-ii

    (c)    P-1-iii, Q-2-ii

    (d)    P-2-ii, Q-1-i

Ans.    (d)

Sol.    C₃ plants that survive solely on C₃ fixation (C₃ plant) tend to thrive in areas where sunlight intensity is moderate, temperatures are moderate, carbon dioxide (CO₂) concentrations are around 200 ppm or higher and ground water is plentiful.

    C₄ plants are native to the tropics and warm temperate zones with high light intensity and high temperature under these conditions, C₄ plants exhibit higher photosynthetic and growth rates due to gains in the water, carbon and nitrogen efficiency uses.

82.    Which one of the following hormones is responsible for mobilizing calcium from the bone and increasing urinary excretion of phosphate ?

    (a)    Calcitonin

    (b)    Angiotensin II

    (c)    Parathormore

    (d)    Vasopessin

Ans.    (c)

Sol.    Parathyroid hormone is secreted from four parathyroid glands, which are small glands in the neck, located behind the thyroid gland. This hormone regulates calcium levels in the blood largely by increasing the levels, when they are too low. It does this through its actions on the kedneys, bones and intestine.

    Parathyroid hormone stimulates the release of calcium from large calcium stores in the bones into the bloodstream. This increases bone destruction and decreases the formation of new bone. This hormone reduces loss of calcium in urine and also, stimulates the production of active vitamin D in the kedneys. Parathyroid hormone indirectly increases calcium absorption from food in the intestine, via its effects in vitamin D metabolism.

    If calcium level rises above set point, then blood calcium level rises.

83.    The sodium-independent iodide/chloride transporter is named as

    (a)    megalin

    (b)    pendrin

    (c)    transthyretin

    (d)    prestin

Ans.    (b)

Sol.    Pendrin was initially identified as a sodium-independent iodine/chlorine exchanger with subsequent studies showing that is also accepts formate and bicarbonate as substrates. It is similar to the band transport protein found in red blood cells (RBCs).

    

84.    Which one of the following is not true in the pocess of acclimatization to high altitude ?

    (a)    Respiratory alkalosis

    (b)    Increased 2, 3-DPG in RBC

    (c)    Rise in pH of cerebrospinal fluid

    (d)    Increased cytochrome oxidase in tissues

Ans.    (c)

Sol.    The major cause of altitude illness is going too high to fast. Given, time, your body adapt to the decrease in oxygen molecules at a specific altitude. This process is known as acclimatization and generally takes 1-3 days at that altitude. The process of acclimatization to high altitude are respiratory alkalosis, increased 2, 3-D PG in RBC and increased cytochrome oxidase in tissues. But not rise in ph of cerebrospinal fluid.

    

85.    In the table below column A lists ligands and column B lists classes of receptors :

    

    Which one is the correct match ?

    (a)    P–1, Q–2, R–3, S–4

    (b)    P–2, Q–3, R–4, S–1

    (c)    P–3, Q–4, R–1, S–2

    (d)    P–4, Q–1, R–2, S–3

Ans.    (a)

Sol.    Serotonin receptors are G-protein coupled receptors involved in a variety of psychiatric disorders. It is binding activates a G-protein which activates an enzymes that generates a specific second messanger and opens ion channel.

    Interferons binding of interferon--causes dimerization of interferon α-receptor.

    Binding causes receptor monomers to dimerize.

    Glycine is an amino acid that has a single hydrogen atom as its side chain. It is one of the proteinogenic acids. Binding changes the conformation of the receptor so that the specific ions flow through it.

    Insulin requires receptor molecules on the cell membrane is due to the large size of the molecules of hormone. Binding leads to activation of intrinsic tyrosine kinase activity. Hence, correct match is (a)-1, (b)-2, (c)-3, (d)-4.

86.    Thyroid hormone (T₃) increases the heart rate and the force of contraction of cardiac muscles. The mechanisms of these effects of T₃ have been explained by a researcher in the following statements :

    1.    T₃ inhibits the expression of gene for -myosin heavy chain and enhances the expression of gene for -myosin heavy chain.

    2.    The expression of gene for Na⁺-Ca⁺⁺ antiporter is enhanced by T₃.

    3.    The sarcoplasmic reticulum Ca⁺⁺ ATPase is increased by T₃.

    4.    T₃ increases ryanodine Ca⁺⁺ channels in the sarcoplasmic reticulum.

    5.    The number of -adrenergic receptors in heart muscles is inhibited by T₃.

    Which one of the following combinations contains both correct explanations ?

    (a)    1 and 2

    (b)    2 and 3

    (c)    3 and 4

    (d)    4 and 5

Ans.    (c)

Sol.    Thyroid hormone (T₃) increases the heart rate and the force of contraction of cardiac muscles. This occurs when The sarcoplasmic reticulum Ca⁺⁺ ATPase is increased by T₃ and T₃ increases ryanodine Ca⁺⁺ channels in the sarcoplasmic reticulum.

87.    Dawn below is an intestinal epithelial cell (IEC) performing the absorption of digested monosaccharides from the dietary carbohydrates ingensted.

    

    Which one of the following combinations of the trannsporter (A, B and C in the figure above) and the transported monosaccharide is correct ?

    (a)    A-GLUT5, B-SGLT1, C-GLUT2

    (b)    A-GLUT2, B-GLUT5, C-SGLT1

    (c)    A-SGLT1, B-GLUT2, C-GLUT5

    (d)    A-SGLT1, B-GLUT5, C-GLUT2

Ans.    (a)

Sol.    Option (a) is correct.

88.    Hormones act be poducing/activating a variety of effectors intracellularly. Below are given a variety of effectors in column 'X' and hormones in column 'Y'

        Column X                Column Y

    P.    Inositol triphosphate (IP₃)    1.    Leptin

    Q.    cGMP            2.    IGF-1

    R.    cAMP            3.    Oxytocin

    S.    Receptor Kinase            4.    Somatostatin

    T.    Associated Kinase            5.    ANF

    Which one of the following combinations of effectors and the specific hormone is correct?

    (a)    P-1 and Q-2

    (b)    Q-4 and P-3

    (c)    R-4 amd S-2

    (d)    T-2 and R-1

Ans.    (c)

Sol.    (a)    Inositol triphosphate abbreviated InSP₃ or IP₃ is an inositol phosphate signaling molecule.

    (b)    cGMP refers to the current good manufacturing practice regulations enforced by the FDA, cFMPs provide for systems that assure proper design, monitoring and control of manufacturing processes and facilities.

    (c)    C-Somatostatin acts by inhibiting the cycle-3', 5'-Adenosine Monophosphate (cAMP)/Protein kinase A pathway, cAMP response element binding protein (cREB) phosphoryaltion and cREB transcription Potency.

    (d)    The insulin like growth factor 1 (IGF-1) receptor is a protein found on the surface of human cells. It is a transmembrane receptor i.e., activated by a hormone called insulin like growth factors (IGF-1) and by a related hormone called IGF-2. It belongs to the large class tyrosine kinase receptors.

    (e)    Associated kinase : The interlinking-1 receptor (IL-R) associated kinase family plays a crucial role in the protective response to pathogens introduced into the human body by inducing acute inflammation followed by additional adaptive immune responses. Hence, correct option is (C).

89.    The ECG recoded by different leads is analysed on the basis of variation of electrical potential at various loci on the surface of the body, and the time scale relation of different waves. After analysing the ECG, following particlas of heart are proposed to be obtained:

    1.    Stoke volumne and cardiac output.

    2.    Volume and pressure changes during cardiac cycle.

    3.    Anatomical orientation of heart.

    4.    Various disturbances in the rhythm and conduction of cardiac excitation.

    5.    The extent, location and progerss of ischemic damage to myocardium.

    Which one of the following combinatiosn represents both incorrect particulars of heart ?

    (a)    1 and 2

    (b)    2 and 3

    (c)    3 and 4

    (d)    4 and 5

Ans.    (a)

Sol.    An ECG lead leads a graphical representation of heart's electrical activity which is calculated by analysing data from several ECG electrodes. After analysing the ECG, the stke volume and cardiac output and volume and pressure changes during cardiac cycle of hear are proposed to be obtained.

90.    Aldosterone increases the reabsoption of Na⁺ from the tubular fluid in the thick ascending limb of loop of Henle and in the distal tble. These effects are explained in the following proposed statements :

    P.    Aldosterone increases the number of Na⁺-Cl^– symporter in the apical membane of principal cells in the early portion of distal tubule.

    Q.    The number of Na⁺ channels (ENaC) is increased in the apical membane of principal cells in the late portion of distal tuble by aldosterone.

    R.    The synthesis of Na⁺, K⁺-ATPase in the basolateral portion of pincipal cells in distal tubule is decreased by the action of aldosterone.

    S.    Aldosterone incerases the reabsorptio of Na⁺ across the apical cell membrane in the thick ascending limb of loop of Henle by decreasing Na⁺, K⁺-ATPase in it.

    Which one of the following combinations represents bothe correct statements ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    P and S

Ans.    (a)

Sol.    Aldosterone increases the reabsorption of Na⁺ from the tubular fluid in the thick ascending limb of loop of Henle and is the distal tubule. It increases the number of Na⁺–Cl^– symporter in the apical membrane of the principal cells in the early portion of distal tubule. The number of Na⁺ channels is increased in the apical membrane of principal cells in the late portion of distal tubule by aldosterone.

91.    Interacting genes which are involved in producing continuous variation in phenotypes in a population are known as/constitute

    (a)    codominant genes

    (b)    pseudogenes

    (c)    alleles

    (d)    QTLs

Ans.    (d)

Sol.    The quantitative trait locus (QTL) is a locus (selection of DNA) that correlates with variation of a quantitative trait in the phenotype of a population of organisms. QTLs are mapped by identifying which molecular markers such as SNPs or AFLPs correlated with an observed trait.

92.    In smmer squash, white colour fruit (W) is dominant over yellow colour (w) and disc-shaped phenotype (D) is dominant over sphere-shaped phenotype (d). Determine the genotype of the parents if the cross between white sphere crossed with white sphere gives 3/4 white sphere and 1/4 yellow sphere.

    (a)    WWDD × wwdd

    (b)    Wwdd × Wwdd

    (c)    WwDd × wwdd

    (d)    wwDD × WWdd

Ans.    (b)

Sol.    W white

    w yellow

    D = disc

    d = sphere

    Wwdd × Wwdd

    Hence, Wwdd × Wwdd

93.    In an organism, a⁺ allele governs gray body colour, while its mutant allele a gives yellow body colour. Further, presence of b⁺ allele gives long and thin hairs while b allele gives rise to short and thick hairs. The alleles a⁺ and b⁺ are dominant over a and b, respectively. An individual with the genotype

    

    has a patch of yellow cells with short and thick hairs. Which one of the following events is most likely to lead to the above ?

    (a)    Non-disjunction of the homologous chomosomes during mitosis.

    (b)    Somatic recombination involving a and b.

    (c)    Translocation occurring in a few somatic cells.

    (d)    Mutation of both a⁺ and b⁺ alleles in the somatic cells.

Ans.    (b)

Sol.    Somatic recombination involving a and b can produce a patch of yellow cells with short and thick hairs.

94.    A mutant mating type mt strain of Chlamydomonas that was resistant to the antibiotic kanamycin (kan^r) and herbicide PPT (ppt^r) was crossed to a wild type mating mt⁺ kan^s ppt^s strain that was sensitive to kanamycin and PPT. Twenty tetrads of the progeny were analyzed for mating type and resistance/sensitivity to kanamycin and PPT. The following observations were made :

    

    The following statements weer made to explain the observations :

    P.    mt and ppt loci are one two different chromosomes.

    Q.    Inheritance of mating type and ppt-resistance/sensitivity are demonstrating cytoplasmic inheritance.

    R.    Inheritance of kanamycin-resistance/sensitivity is demonstrating nuclear inheritance.

    S.    Nuclear inheritance is being demonstrated by mating type and ppt-resistance/sensitivity analysis.

    Which one of the combinations of above statements is correct ?

    (a)    P and Q

    (b)    P and S

    (c)    Q and R

    (d)    R and S

Ans.    (b)

Sol.    

    Thus, we observed that Mt and ppt loci are on two different chromosomes and nuclear inheritance is being demonstrated by mating type and ppt resistance analysis.

95.    A pedigree shown below depicts that the individual I-1 is heterozygous for a dominant disease allele D and for molecular markers M1/M2. The paternal molecular markes in the progeny individuals are indicated in the pedigree.

    

    The following statements may be drawn from the above pedigree :

    P.    The two loci D/d and M1/M2 appears to be linked.

    Q.    The recombination frequency between the two loci is 20%.

    R.    If LOD score comes out to be 3, then it ensures that the two loci are inndependently assorting.

    S.    A LOD score <1 would have ensured that the two loci aer linked.

    Which combination of the above statements can corectly interpret the depicted pedigree ?

    (a)    R and S

    (b)    Only R

    (c)    P and Q

    (d)    Only S

Ans.    (c)

Sol.    The two loci D/d and M1|M2 appears to be linked and recombination frequency between the two loci is 20%.

    

96.    An individual is having a paracentric inversion (denoted by the region f-e-d, marked by arrows) in homozygous condition.

    

    The meiotic consequences of inversion can be

    P.    generation of an acentric and a dicentric chromosome.

    Q.    the recombinants will hae long deletion or duplication and may be lethal.

    R.    the inversion will suppress crossing over.

    S.    All gametes will have complete genome and will survive normally.

    Which of the above statement or their combinations will explain the meiotic consequences of the given inversion logically ?

    (a)    P, Q and R

    (b)    P and Q

    (c)    Q and R

    (d)    Only S

Ans.    (d)

Sol.    During meiosis, the chromosome with paracentric inversion at the points mentioned in diagram would produce All gametes will have complete genome and will survive normally.

97.    In the following pedigree three STR loci A, B and C are linked on the long arm of the X-chromosome in the order centromere-A-B-C-telomere. Further in the table, the STR alleles present in each individual is indicated.

    

    

    Based on the above, X-chromosome(s) in which of the following individuals are recombinant ?

    [Hint : X-chromosome in males will help identify the phase of the alleles]

    (a)    II-1, III-1 and III-2

    (b)    II-2, III-1 and III-2

    (c)    III-1 and III-3

    (d)    III-2 and III-4

Ans.    (c)

Sol.    By observing the table given in question III-1 and III-3 would have recombinant X chromosome.

98.    The individuals considered in this question are having two haploid sets of autosomes and no Y-chromosome. The X : A ratio of the individuals, the type of organisms chosen, their primary sex and number of Barr bodies expected in their cells are shown in the table below :

    

    Select the option below with all correct matches :

    (a)    P-1-ii-a, Q-1-ii-d, P-2-i-a, Q-2-iii-a

    (b)    P-1-i-a, Q-1-ii-c, P-2-ii-a, Q-2-i-a

    (c)    P-1-ii-c, Q-1-i-d, P-2-i-c, Q-2-ii-b

    (d)    P-1-ii-a, Q-1-ii-d, P-2-iii-a, Q-2-i-a

Ans.    (a)

Sol.    The correct match will be P-1-ii-a, Q-1-ii-d, P-2-i-a, Q-2-iii-a.

99.    A disease-resistant plant was crossed with a susceptible plant and the resultant F1 plants were disease resistant. The F1 plant was selfed and the F2 individuals were analyzed for qualitative and quantitative disease resistance. The following statements were hypothesized :

    P.    Qualitative resistance follows Mendelian ratio.

    Q.    In the F2 individuals demonstrating qualitative resistance, "resistance" is dominant.

    R.    Quantitative resistance is always monogenic.

    S.    Qualitative resistance can be polygenic.

    Which one of the following combination of statements is correct ?

    (a)    P, R and S

    (b)    P, Q and R

    (c)    P, Q and S

    (d)    Q, R and S

Ans.    (c)

Sol.    The disease resistance plant crossed with a plant and resultant F1 plants.

    Plant F1 (selfed) and plant F2 (individuals) are :

    Qualatative and quantitative diesase resistance. The hyppothesized statements are :

    Qualitative resistance Mendelian ratio

    Qualitative Generally monogenic

    Quantitative Generally polygenic

    Quantitative resistance monogenic

    Qualitative resistance polygenic

    Hence correct answer is (c).

100.    Which one of the following show complete metamorphosis in all three orders ?

    (a)    Coleopterans, Dipterans and Hymenopterans.

    (b)    Coleopterans, Hymenopterans and Orthopterans.

    (c)    Dipterans, Lepidopterans and Hemipterans.

    (d)    Hymenopterans, Lepidopterans and Orthopterans.

Ans.    (a)

Sol.    Coleopterans, Dipterans and Hymenopterans show the metamorphosis in all three orders.

101.    Which one of the following plants has this combination of key plant traits : sporophyte dominant in the lifecycle, vascular tissue, lack of seeds ?

    (a)    Mosses

    (b)    Ferns

    (c)    Cycads

    (d)    Monocots

Ans.    (b)

Sol.    The life cycle of seedless vascular plants alternates between adiploid sporophyte and haploid gametophyte phase. Seedless vascular plants reproduce through unicellular, haploid spores instead of seeds; the light weight spores allow for easy disperson in the wind. These plants require water for sperm motility during reproduction and thus, are often found in moist environments.

    Life cycle of a fern : This life cycle of a fern shows alternation of generations with a dominant sporophyte stage.

    

102.    Cnidarians are

    (a)    triploblastic animals with bilateral symmetry.

    (b)    diploblastic animals with medusa as one of the basic body forms.

    (c)    monoblastic organisms with tube feet.

    (d)    asymmetric organisms with tentacles containing poison glands.

Ans.    (b)

Sol.    Cnidarians are phylum under kingdom animalia containing over 11000 species of aquatic animals found both in freshwater and marine environment, they are pre dominantly marine. Their distinguishing feature is cnidoytes, specialized cells that they use mainly for capturing prey. They are diploblastic animals with medusa as one the basic body forms.

103.    Which one of the following was recently reported to be the first mammal to have become extinct as a result of climate change ?

    (a)    Bramble Cay melomys–Melomys rubicola

    (b)    Gangetic river dolphin–Platanista gangetica

    (c)    Malaga giant rat–Hypogeomys antimena

    (d)    Tapanuli orangutan–Pongo tapanliensis

Ans.    (a)

Sol.    The Melomys rubicola, also referred to as the Bramble cay Mosaic tailed rat was found in Bramble cay, a small vegetable coral key at Australia's extreme north. The rodent, which built furrows in herd fields among strandile plants has been feared to be extinct for sometime now. The government of Australia's Queensland province reported the species to be extinct in June 2016. It was place in the IUCN (International Union for Conservation of Nature) Red list of threatened species.

104.    In a lake, reducing the population of a fish which feeds on plankton was followed by a decline in the rate of primary prodctivity. This is consistent with which one of the following hypotheses regarding the regulation of primary productivity ?

    (a)    Bottom-up control

    (b)    Eutrophication

    (c)    Top-down control

    (d)    Trophic pyramid

Ans.    (c)

Sol.    The top-down and bottom up controls were investigated by assessing the impact of omnivorous plantivorous fish.

105.    An interaction where the actor and the recipient both suffer a cost is referred to as

    (a)    Altuism

    (b)    Cooperation

    (c)    Mutualism

    (d)    Spite

Ans.    (d)

Sol.    When both the actor and the recipient suffer as a result of an interaction (–, –) it is called spite. Conventional wisdom has it that to be spiteful is the prerogative of human alone. It is true that there is no unequivocal example of interaction between animals where both actor and recipient clearly suffer.

106.    During the exponential phase of growth, if N_o, N_t, and n define the initial population number, population number as time t and the number of generations in time t, respectively, then

    (a)    N_t = N_o × 2n

    (b)    N_o = N_t/2

    (c)    N_t/N_o = 2ⁿ

    (d)    N_o/N_t = 2ⁿ

Ans.    (c)

Sol.    We have, N = No × 2ⁿ

    

    N_t = N₀ × 2ⁿ

    

107.    Several plants produce metabolites with important medicinal properties and have been extensively used in traditional medicine across the world. Many of these compounds can now be purified or synthesized and are used in modern medicine. Given below is a list of metabolites, their plant source and medicinal use :

    

    Which one of of the following options is the most appropriate match of the compound with its plant soruce and use ?

    (a)    P-3-ii, Q-1-iv, R-4-i, S-2-iii

    (b)    P-4-i, Q-3-ii, R-2-iii, S-1-iv

    (c)    P-2-iv, Q-3-iii, R-1-ii, S-4-i

    (d)    P-3-iii, Q-4-i, R-2-iv, S-1-ii

Ans.    (d)

Sol.    Digoxin – Digitalis purponea – Anti malerial

    Salicin – Artemisia annua – Narcotic analgesic

    Morphine – Willow tree – Aspirin

    Artemignin – Papaver somniferum – Cardia ailment

108.    Match the following taxa with the genus of the microorganism

        Taxa                Genus

    P.    Ascomycota            1.    Rhizopus

    Q.    Basidiomycota            2.    Erysiphe

    R.    Zygomycota            3.    Pythium

    S.    Oomycota            4.    Ustilago

    (a)    P–2, Q–4, R–1, S–3

    (b)    P–2, Q–3, R–2, S–4

    (c)    P–3, Q–4, R–3, S–1

    (d)    P–1, Q–2, R–4, S–2

Ans.    (a)

Sol.    A.    Ascomycota – Erysiphe

    B.    Basidiomycota – Ustilago

    C.    Zygomycota – Rhizopus

    D.    Oomyceta – Pythium

    Hence (a)-2, (b)-4, (c)-1, (d)-3

109.    Match the following plant diseases with the name of pathogen associated with the disease:

        Disease                Pathogen

    P.    Powdery mildew            1.    Erwinia amylovora

    Q.    Rice blase            2.    Pseudomonas syringae pv. syringae

    R.    Bacterial canke            3.    Magnaporthe oryzae

    S.    Fire blight            4.    Erysiphe cichoracearum

    (a)    P–2, Q–3, R–1, S–4

    (b)    P–1, Q–4, R–2, S–3

    (c)    P–4, Q–3, R–2, S–1

    (d)    P–3, Q–2, R–4, S–1

Ans.    (c)

Sol.    (a)    Erysiphe cichoracearum is a fungal plant pathogen that caused powdery mildew disease.

    (b)    Rice blast caused by fungus magnaporthe oryzae.

    (c)    The bacterium that causes canker, pseudomonas syringae.

    (d)    Fire blight, also written fireblight is a contagious disease affecting apples, pears and some other members of family rosacease. It is causedb y Erwinia amylovora.

    Hence, correct option is (C).

110.    Given below are certain adaptations which are seen in various groups of animals :

    1.    Ovipary

    2.    Streamlined body

    3.    Pouch for carrying eggs

    4.    Porous egg shell

    5.    Breast bone as large keel

    6.    Webbed feet

    7.    Laterally compressed coccygeal bone

    8.    Unidirectional pulmonary system to provide large quantities of oxygen

    9.    Barbules or hooklets on the vanes of each feather

    Which combination of the above adaptations facilitate bird flight ?

    (a)    2, 5, 8, 9

    (b)    1, 2, 3, 7

    (c)    4, 6, 7, 9

    (d)    2, 4, 5, 6

Ans.    (a)

Sol.    Some adaptations are seen in the various groups of animals :

    Streamline body,

    Breast bone are large keel,

    Unidirectional pulmonary system to provide large quantities of oxygen

    Barbules of hooklets on the vanes of each feather.

    Hence, correct answer is (a).

111.    Which of the following set of conditions will qualify a species to be considered as endangered (EN) as per IUCN criteria ?

    (a)    80% reduction in population size, <100 km² area of occupancy, at least 50% estimated extinction risk in five generations.

    (b)    <2,500 individuals with declining population, <250 mature individuals, at least 20% estimated extinction risk in 10 years.

    (c)    <10,000 individuals with declining population, <1000 matre individuals, at least 10% estimated extinction risk in 10 years.

    (d)    75% reduction in population size, <500 km² area of occpancy, at least 20% estimated extinction risk in five generations.

Ans.    (d)

Sol.    A species, whose numbers are so small that the species is at risk of extinction. To be defined as endangered, a species mus meet any of the following criteria.

        Population reduction : 50-70% population decline

        Geographic range

        Extent of occurrence : <5000 km²

        Area of occupancy : <500 km²

        Population size : <2500 mature individuals

        Extinction probability (in the wild) : at least 20% within 20 years or 5 generations

        Example : Red panda, Snow leopard, Bengal Tiger and One horned rhinoceros and Black buck.

112.    In the graph below, large boxes (deonoted by P, Q, R, S) represent a region, whereas smaller boxes represent habitats. Labels S₁, S₂, ...., above each small pox represent species present in the given habitat denoted by that box.

    P.    

    Q.    

    R.    

    S.    

    Given the above graphs, choose the option which correctly depicts the regions which show maximum and maximum divesity, respectively.

    (a)    Q and S

    (b)    P and R

    (c)    S and P

    (d)    Q and R

Ans.    (a)

Sol.    

    

    Q shows the maximum -diversity and S shows the maximum -diversity.

    Hence, A is correct answer.

113.    Which of the following is not a mechanism fo species coexistence ?

    (a)    Niche differentiation

    (b)    Niche complementarity

    (c)    Niche overlap

    (d)    Amount of limiting resorces is greater than the numbe of species

Ans.    (c)

Sol.    Niche overlap occurs when two organismic units use the same resources or other environmental variables. So it is not mechanism for species co-existence.

    The term niche differentiation, as it applies to the field of ecology, refers to the process by which competing species use the ebvironment differently in a way that helps the coexists. Niche complementarity refers to the concept in which co-existence species can overlap in fundamental niches due to using different forms of resources.

    Hence, option (c) is correct.

114.    Which of the following represents exponential growth in populations ?

    (a)

    (b)

    (c)

    (d)

Ans.    (b)

Sol.    

    

    

    

    

115.    A smaller number (approximately 10) of mice are introduced into an uninhabited island. Their population grows exponentially initially and after 3 years, reaches a population size of 520 after which the population becomes stable. At what point would you expect their population to attain their highest growth rate ?

    (a)    When the mice population was fist introduced.

    (b)    When the population size is 260.

    (c)    Their population growth rate remains constant throughout.

    (d)    When the population size reaches 520.

Ans.    (b)

Sol.    Maximum population growth rate occurs when N = K/2; here K = carrying capacity.

    So, maximum population size = K/2 = 520/2 = 260.

116.    The frequency of homozygotes in a diploid population is 0.68. Assuming that the population is in Hardy-Weinbeg equilibrium, the frequencies of the two alleles are

    (a)    0.1 and 0.9

    (b)    0.2 and 0.8

    (c)    0.4 and 0.6

    (d)    0.5 and 0.5

Ans.    (b)

Sol.    We have,    p² + q²    =    0.68

        2pq    =    2 × 0.2 × 0.8 = 0.32

    Homozygote

    

        q    =    1 – 0.8 = 0.2

    A.    0.1 and 0.9    =    (0.1)² + (0.9)²

            =    0.01 + 0.81 = 0.82

    B.    0.2 and 0.8    =    (0.2)² + (0.8)²

            =    0.04 + 0.64 = 0.68

    C.    0.4 and 0.6    =    (0.4)² + (0.6)²

            =    0.16 + 0.36 = 0.50

    D.    0.5 and 0.5    =    (0.5)² + (0.5)²

            =    0.25 + 0.25 = 0.50

    Hence, correct option is (b).

117.    The ratio of variance in male mating success (V_m) to variance in female mating success (V_f) is strongly male biased (V_m > V_f) in species P, strongly female biased in species Q (V_f > V_m) and similar in species R (V_m = V_f). All else being equal, which one of the following matches between species and mating systems is most likely ?

    (a)    P-monogamy, Q-polyandry, R-polygyny

    (b)    P-polyandry, Q-polygyny, R-monogamy

    (c)    P-polygyny, Q-polyandry, R-monogamy

    (d)    P-monogamy, Q-polygyny, R-polyandry

Ans.    (c)

Sol.    

118.    The phylogeny below shows evolutionary between 9 extant bird species and whether they display red or blue plumage. Based on the above phylogeny and the distribution of red and blue character states among the extant species, and using the principle of parsimony, which of the folowign is the correct inferences about the plumage colour of the ancestor at the root P ?

    

    (a)    Ancestral state at P is blue.

    (b)    Ancestral state at P is red.

    (c)    Ancestral state at P is more likely to be red than blue.

    (d)    Ancestral state at P is eqally likely to be red or blue.

Ans.    (a)

Sol.    In general, parsimony is the principle that the simplest explanation that can explain the data is to be preferred. In the analysis of phylogeny, parsimony means that a hypothesis of relationships that requires the smallest number of character changes is most likely to be correct. In this phylogeny, ancestral state of P is blue because two last of its least diverged species show blue plumage. Red colour is through evolution.

119.    The top panel (graphs P-R) represents trends of number of sperms produced per mating season with respect to number of mates, while the bottom panel (graphs 1-4) represents trends of time invested in paternal care with respect to number of mates in birds.

    

    

    Select the correct trend from each panel.

    (a)    R, 4

    (b)    P, 2

    (c)    Q, 3

    (d)    P, 1

Ans.    (b)

Sol.    Sperm count more it will not care.

    Parental care ×

    Hence, correct trend from each panel is a, (ii). i.e., correct option is (B).

120.    Which one of the following phylogenies best represents the evolutionary relationship among whales, dolphins, seals, deer and dogs ?

    (a)

    (b)

    (c)

    (d)

Ans.    (b)

Sol.    B phylogenetic tree shows the correct evolutionary relationship among whales, dolphins, seals,deer and dogs .Marine mammals in the cetacean family include whales, dolphins, and porpoises. Seals are a group of species that are very closely related to dogs, so much so that they can catch canine distemper.

121.    Match the geological time period with the extinction of diversification events associated with them :

        Geological time                Event

    P.    Cenozoic            1.    Angiosperm diversification

    Q.    Cretaceous            2.    Modern fauna diversification (bivalves,

                        gastropods, bryozoans, malacostracan

                        crustaceans)

    R.    Paleozoic            3.    Megafauna extinction

    S.    Quatenary            4.    Mammal diversification

    (a)    P–2, Q–1, R–3, S–4

    (b)    P–4, Q–1, R–2, S–3

    (c)    P–2, Q–3, R–1, S–4

    (d)    P–4, Q–3, R–1, S–2

Ans.    (b)

Sol.    Cenozoic, after the dinosaurs became extinct, the number and diversity of mammals exploded.

    Angiosperm diversification and cretaceous floristic trends a comparison of polynofloras and leaf macrofloras.

    Palezoic fauna fauna II is known as paleozoic described as a Brachipod-rich assemblage account for most of the fossils appearing in the great ordovician.

    Quaternary period saw the extinctions of numerous predominantly megafaunal species which resulted in a collapse in faunal density and diversity and the extinction of key ecological strata across th globe.

    Hence, correct option is (b).

122.    

    Which among the above sets of conditions aer best sulted form mimicry to be successful?

    (a)    Condition A

    (b)    Condition B

    (c)    Condition C

    (d)    Condition D

Ans.    (b)

Sol.    Mimicry evolves if a receiver (such as a predator) perceives the similarity between a mimic (the organism that has a resemblance) and a model (the organism it resembles) and as a result changes its behaviour in a way that provides a selective advantage to the mimic. For a mimicry to be successful, model should be abundant than mimic and there should be no resource overlap.

123.    Agrobacterium tumefaciens is frequently used as a vector to create transgenic plants. Under laboratory conditions Agrobacterium-mediated plant transformation does not require

    (a)    host plant genes

    (b)    bacterial type IV secretion system

    (c)    vir genes

    (d)    opine catabolism genes

Ans.    (d)

Sol.    Agrobacterium tumefaciens is frequently used as a vector to create transgenic plants. Under laboratory conditions agrobacterium mediated plant transformation requires host plant genes, bacterial type VI secretion system and Vir genes but does not required opine catabolism genes.

    

    Hence, correct option is (D).

124.    The T-DNA region of the Ti plasmid of Agrobacterium tumefaciens harbous two genes : X and Y. Mutation of gene 'X' produces a rooty tumour while mutation of gene 'Y' produces shoots in the tumor. Based on the above information, which one of the following statements is correct ?

    (a)    Gene 'X' encodes auxins and gene 'Y' encodes cytokinins.

    (b)    Gene 'X' encodes cytokiniins and gene 'Y' encodes auxins.

    (c)    Gene 'X' and gene 'Y' both encode auxins.

    (d)    Gene 'X' encodes opines while gene 'Y' encodes cytokinins.

Ans.    (b)

Sol.    Both auxin and cytokinin have been known for a long time to act either synegistically or antagonistically to control several significant development process, such as formation and maintenance or meristem.

    The frameworks for the complicated interaction of auxin and cytokinin hormones in the control of roots apical meristems and shoot apical meristem formation as well as their roles in vitro organ regeneration are the major focus. Hence, Gene X encodes cytokinins and gene-Y encodes auxins.

125.    The cells of inner cell mass of a blastocyst stage mammalian embryo are

    (a)    totipotent

    (b)    pluripotent

    (c)    multipotent

    (d)    unipotent

Ans.    (b)

Sol.    Human embryonic stem (ES) cells are pluripotenet cell lines that have been derived from the Inner Cell mass (ICM) of blastocyst stage embryos. They are characterized by their ability to be propagated indefinitely in culture as undifferentiated cells with a normal karyotype and can be induced so differentiate in vitro various cell types. So, the pluripotent cells of inner cell mass of blastocyst stage mammalian embryo.

    Totaipotent whole organisms

    Pluripotent all cell

    Multipotent multiple

    Unipotent single

126.    What is the probability of getting a sum of 9 from simultaneously throwing two dice ?

    (a)    1/6

    (b)    1/8

    (c)    1/9

    (d)    1/12

Ans.    (c)

Sol.    We have 6 × 6 = 36

    p² + q² + 2pq

    

    

127.    In electron microscopy, to detect specific macromilecule or structure such as spindle poly body (SPB), the frequently used procedure is to couple secondary antibody with

    (a)    Alexa 568

    (b)    Cy5

    (c)    Gold particle

    (d)    Osmium tetraoxide

Ans.    (c)

Sol.    The electron microscopy product line is extensive and includes tungeston filaments, Lab6 and CeB6 cathodes, grids, silicon nitride films, adhesives and mounts as well as SEM cryo-preparation systems, sample coaters, carbon evaporators, freeze driers and critical point driers. In the electron microscopy to detect specific macro molecule such as spindle pole body, the frequently used procedure is to couple secondary antibody with gold particle.

128.    If you inject a mouse with radioactie material of current activity of 256 Bq, what will be the activity after completion of 6 half-lives ?

    (a)    4 Bq

    (b)    8 Bq

    (c)    16 Bq

    (d)    24 Bq

Ans.    (a)

Sol.    

    Hence, the activity after completion of 6 half-lives will be 4 Bq.

129.    A researcher samples n individals randomly from a population of blackbuck and identifies their sex. The number of females in the sample follows

    (a)    an exponential distribution

    (b)    a binomial distribution

    (c)    a poisson distribution

    (d)    a normal distribution

Ans.    (b)

Sol.    The binomial distribution model gives us the probabilities of the number of females (F), in an independent radom sample of size n. The value of F can be anywhere between zero (no femal in the sample) and n (all sample female).

130.    Readers of histone modifications include

    (a)    SUN domain proteins

    (b)    BAG domain proteins

    (c)    PAS domain proteins

    (d)    TUDOR domain proteins

Ans.    (d)

Sol.    The TUDOR domain is methyl-lysine and methyl-argining binding domain presen in proteins involved in cellular functions as DNA transcription, RNA metabolism, gene silencing the transmission of epigenetic post-translation modifications and the maintenance of genomic stability. We review the different forms TUDOR Single Tudor, tandem Tudor, hybrid Tudor, extended Tudor and subtrate binding modes, including methylation state specificity of the Tuder domain.

    

131.    Which one of the following does not use RNA-sequencing ?

    (a)    Mapping transcription initiation sites.

    (b)    Long non-coding RNA profiling.

    (c)    Alternatie polyadenylation profiling.

    (d)    Mammalian epigenome sequencing.

Ans.    (d)

Sol.    RNA-sequencing is a technique that can examine the quantity and sequence of RNA is a sample using next generation sequencing (NGS). It analyzes the transcriptone of gene expression patterns encoded our RNA.

    Technological advances in the sequencing field support in depth characterization of the transcriptome. Here, genome wide RNA sequencing methods used to investigate specific aspects of gene expression and its regulation. The tag based methods for studing transcription alternative initiation and polyadenylation events than gun methods for detection of alternative splicing, full-length RAN sequencing for determination of complete transcript structures and targeted methods for studying the process of transcription and translation. With the ensemble of technologies available, it is transcriptome complexity and the regulation of transcript diversily.

    Therefore mammalian epigenome sequencing is not used in RNA sequencing.

132.    Which one of the following statements related to molecular cloning procedures in incorrect ?

    (a)    5' overhangs of restricted DNA fragments can be blunt-ended by Klenow polymerase but no by DNaseI.

    (b)    A DNA fragment obtained as an XhoI fragment (CTCGAG) may be ligated at the SalI site (GTCGAC) in a vector.

    (c)    To prevent self-ligation of a vector digested with KpnI (GGTACC), alkaline phosphatase enzyme is used to remove 3'-PO₄ groups from the ends of fragments.

    (d)    -complementation/blue-white screening may produce blue coloured recombinant colonies (containing cloned fragments) in case of translational fusion with the -galactosidase gene.

Ans.    (c)

Sol.    Since Klenow fragment is DNA polymerase with nick filling activity, so blunt ends can be filled by it but not by nucleases. As fragments of XhoI and SalI would be compatible to each other they can be ligated, -complementation results in blue colonies (in an unusual case, usually recombinants are white), when galactosides gene is fused with gene of interest.

    However, statement "To prevent self-ligation of a vector digested with KpnI (GGTACC), alkaline phosphatase enzyme is used to remove 3'-PO₄ groups from the ends of fragments" is incorrect, as PO₄ is present on 5' end which need to be removed to prevent self-ligation.

133.    The following statements are made regarding developing a transgenic mouse :

    1.    The transgenic mouse thus born will be a homozygous transgenic animal and can be maintained by crossing with another transgenic animal.

    2.    The fertilized transgenic eggs are allowed to develop in vitro.

    3.    The desired gene is preferably micoinjected in male pronucleus after sperm entry in oocyte.

    4.    For best efficiency, the desired gene is always microinjected in the male gametes and then they aer allowed to fertilize the female gametes.

    5.    Blastocyst stage embryos are transferred to the uteus of hormonally prepaed mother.

    6.    The fertilized eggs are collected from specific starin of mouse.

    7.    The female mouse of specific strain is superovulated, oocytes are collected and allowed to fertilize in vitro.

    Choose the combination of statements arranged in the correct sequence for developing transgenic mouse.

    (a)    7 3 5

    (b)    6 3 2 1

    (c)    7 4 1

    (d)    4 6 2 1

Ans.    (a)

Sol.    In short, generating transgenic mice involves five basic steps: purification of transgenic construct, harvesting donor zygotes, microinjection of transgenic construct, implantation of microinjected zygotes into the pseudo-pregnant recipient mice . So, the correct sequence to develop transgenic mice would be 7-3-5.

134.    Given below are a few statements on use of plant breeding to develop improved varieties of a crop that :

    P.    Genotypic/phenotypic variation in the desired trait should be available in the germplasm resources of the crop plant.

    Q.    Availability of molecular markers linked to the trait of interest would decelerate the process of trait introgression into desired varieties.

    R.    Breeding procedures to improve plant varieties are generally more successful among sexually compatible species as compared to sexually incompatible species.

    S.    Co-dominant molecular markers cannot be used for selection of plants with the desired trait.

    Which of the above statements(s) is/are incorrect ?

    (a)    P and R

    (b)    Q only

    (c)    R and S

    (d)    Q and S

Ans.    (d)

Sol.    Marker-assisted selection can improve the efficiency and accuracy of conventional plant breeding in many crops. Co-dominant markers are particularly powerful in selecting the desirable traits without missing heterozygous plants because they can distinguish between homozygous and heterozygous plants.

135.    Given below are various types of molecular markers in column A and properties of these markers in column B.

        Column A        Column B

    P.    RFLP    1.    Single locus

    Q.    SSR    2.    Multi-allelic

    R.    AFLP    3.    Co-dominant

    S.    RAPD    4.    Single-allelic

            5.    Multi-locus

            6.    Dominant

    Which one of the options given below correctly matches the molecular markers with their properties ?

    (a)    P–6, Q–1, R–2, S–5

    (b)    P–5, Q–2, R–4, S–3

    (c)    P–1, Q–2, R–5, S–6

    (d)    P–2, Q–3, R–1, S–2

Ans.    (c)

Sol.    RFLP        It is a type of polymorphism that results from variation in the DNA sequences recognized by restriction enzymes. RFLP analysis of forensic enzymes. RFLP analysis of forensic DNA sample with single locus VNTR genetic markers.

    SSR        Simple sequence repeats are stretches containing repeats of short units of nucleotides. SSR marker can be used to differentiate homozygotic and heterozygotic alleles. It is multi-allelic

    AFLP     It is technique used to detect polymorphisms in DNA, when no information about the genome is known. It is multi-locus.

    RAPD        Nearly all RAPD markers are dominant i.e., it is not distinguished whether a DNA segment is amplified from a locus i.e., heterozygous.

136.    Given below are schematic representations of the T-DNA regions of four constructs that are to be used for Agrobacterium-mediated transformation to silence an endogenous plant gene represented as 'XYFP' that is expressed constitutively in the plant.

    

    

    

    M : Selectable marker gene expression cassette.

    LB : Left Border

    RB : Right Border

    Which of the fou constructs depicted above could be used to silence the target gene 'XYFP' ?

    (a)    P and Q only

    (b)    Q and S only

    (c)    P and R only

    (d)    R and S only

Ans.    (d)

Sol.    

137.    Analysis of a homotetrameric protein and a double stranded DNA (that had been incubated in standard buffer) on native gels revealed that they migrated due to their physical states (tetrameric nature of the protein and double stranded natue of the DNA). Following hypotheses were made for the effect of adding high salt to the incubation mix and subsequent analysis on native gels.

    P.    The protein would migrate as a homotetramer and DNA in double standed form.

    Q.    The protein would migate as a monomer but DNA in double standed form.

    R.    The protein would migarte as a homotetramer but the DNA in single stranded form.

    S.    The protein would migrate as a monomer and the DNA in single stranded form.

    Which of the following combiantion of hypotheses is most likely ?

    (a)    P and Q

    (b)    Q and R

    (c)    R and S

    (d)    P and S

Ans.    (a)

Sol.    

    The homotetrameric protein analysis and double strandard DNA on native gels revealed that they migrated true to their physical states. The protein would migrate as a homotetramer and DNA in double standard form and the protein would migrate as a monomer but DNA in double strandard forms were made for the effect high salt to the incubation mix and subsequent analysis on native gels.

138.    Female fiddler crabs prefer male fiddler crabs with larger claws over males with smaller claws. If the selection pressure exerted is strong resulting in a skewed distribution of clas size, which of the following statements is true abot the population's mean, median and mode ?

    (a)    Mean > Median > Mode

    (b)    Mean < Median < Mode

    (c)    Mean = Mode < Median

    (d)    Mean = Median = Mode

Ans.    (b)

Sol.    The female like larger claws. They likely indicate a male that's big. Male can offer her a big burrow, because she goes into that to incubate her eggs. Also, crustaceans continue to grow for majority of their lives, so a bigger male is older, which means that he is a survivor so he'd be a good one to mate with, when the selection pressure exerted in strong resulting in a skewed distribution of claw sisze. i.g., True statement are Mean < Median < Mode.

139.    In an experiment, a 1 kb fragment with a single BamHI site (as shown below in figure 'A') is to be cloned in the SmaI (CCCGGG) site of a clonign vector of 3 kb length (figure 'B'). None of the other enzymes of the multiple cloning site are present in the fragment to be cloned.

    

    Based on the information given above, a seriaes of digestion were set up for the potential clones and their fragment profiles are given below :

    A.    BamHI    :    200 bp + 3.8 kb

    B.    BamHI    :    800 bp + 3.2 kb

    C.    HindIII + EcoRI    :    ~1 kb + ~3 kb

    D.    XhoI + BamHI    :    ~200 bp + ~800 bp + ~3 kb

    Which one of the above digestion profiles confirms successful cloning of the fragment in the vector in an orientation wherein the 5' end of the cloned fragment is towards 'P' ?

    (a)    A only

    (b)    B only

    (c)    A and C

    (d)    C and D

Ans.    (b)

Sol.    

    Bam HI : 800 bp + 3.2 kb is the digestion profiles confirms successful cloning of the fragment in vector in an orientation where the 5' end of the cloned fragment is towards P'.

140.    Given below are some physiochemical properties (column X) and their manifestations (column Y).

        Column X                Column Y

    P.    Pauling electronegativity    1.    Charge separation

    Q.    Isolated -orbital overlap    2.    Solvation of atoms

    R.    Aromaticity            3.    Restricted rotation

    S.    Dielectric constant        4.    Planarity of molecules

    Which one of the following is the most appropirate match ?

    (a)    P–1, Q–4, R–2, S–3

    (b)    P–3, Q–2, R–4, S–1

    (c)    P–2, Q–3, R–4, S–1

    (d)    P–4, Q–2, R–3, S–1

Ans.    (c)

Sol.    Pauling electronegativity is the solvation of atoms.

    Isolated -orbital overlap prevents rotation because of the electron overlap both above the below the plane of the atoms.

    Aromaticity is a property of conjugated cyclo alkenes in which the stabilization of the molecule is enhanced due to the ability of the electron in the π-orbitals to decolize. This act as a framework to create a planar moelcules.

    Dielectric constant, when the molecules of a dielectric are placed in the electric field their negatively charged electrons separates slightly from their positively charged cores. With this separation, referred to as polarization, the molecules acquire an electric dipole moment.

    Correct match (a)-2, (b)-3, (c)-4, (d)-1.

    Hence, correct option is (C).

141.    Absorption spectra of L-tyrosine in acidic (continuous line) and basic (dotted line) medium was estimated and plotted on a graph as depicted below :

    

    Following interpretations were made :

    P.    Change in the pH from acidic to basic results in shift in the lowest energy absorption maximm and decrease in the molar absorptivity.

    Q.    Shifting of the absorption band to longer wavelength signifies a shift to lower energy, also known as red shift.

    R.    Shifting of the absorption band to shorter wavelength signifies a shift to highe energy, also known as blue shift.

    S.    Wavelength shift is always accompanied by change in intensity of the absorption band.

    Select the combination with correct interpretations.

    (a)    P and Q

    (b)    P and R

    (c)    Q and R

    (d)    Q and S

Ans.    (c)

Sol.    

142.    A mixed cell population was stained with two antibodies, one specific fo cell surface antigen A and the other specific for cell surface antigen B. Anti-A antibody was labelled with fluorescein and anti-B was labelled with rhodamine. The cell population was then analysed for the presence of antigens by flow cytometry. Which one of the following is the correct outcome for this cell population ?

    (a)

    (b)

    (c)

    (d)

Ans.    (c)

Sol.    The results of flow cytometry are obtained in the form of a scatter plot graph, having one horizontal axis (x-axis) and one vertical axis (y-axis) that divides the whole graph into four quadrants that can be named as upper right, upper left, lower left and lower right quadrant (also called first, second, third and fourth quadrants respectively).

    Upper right quadrant shows positive results for both x and y axis, lower left quadrant shows negative results for both x and y axis, upper left quadrant shows positive result for y axis and negative result for x axis, while lower right quadrant shows negative results for y axis and positive result for x axis.

    Fluorescein produces green fluorescence and was conjugated to antibody binding with Antigen A, while rhodamine fluorescence red and was conjugated to antibodies binding with antigen B, hence, lower left quadrant will be representing A^–B^– cells, lower right quadrant will be representing A⁺B^– cells, upper left quadrant will represent A^–B⁺ cells and upper right quadrant would represent A⁺B⁺ cells.

143.    Given below are two sets of terms related to various methods used in biological science.

        Column A                Column B

    P.    RACE            1.    DNA-protein inteactions

    Q.    South-Western blotting    2.    FAM

    R.    Recrusive PCR            3.    Determining the ends of mRNA

    S.    TaqMan            4.    Construction of synthetic DNA

    Which one of the following options correctly matches terms of column A and column B ?

    (a)    P–4, Q–3, R–1, S–2

    (b)    P–3, Q–1, R–4, S–2

    (c)    P–2, Q–4, R–1, S–3

    (d)    P–2, Q–1, R–4, S–3

Ans.    (b)

Sol.    RACE (Rapid amplification of cDNA ends) is a technique used in molecular biology to obtain the full length sequence of an RNA transcript found within a cell. It is determining the ends of mRNA.

    South-western blotting is a technique used to study DNA protein interactions. This method detects specific DNA binding proteins by incubating radio labelled DNA with a gel blot, washing and visualizing through autoradiography.

    Recursive PCR cna be used to construct long, double stranded DNA from appropriate collections of long DNA oligonucleotides in a process. It is construction of synthetic DNA.

    TaqMan gene expression assays are used for quantative real time PCR analysis of gene expression and consist of a pair of unlabelled PCR primers and a TaqMan probe with a dye label (FAM) on the 5' end and a minor groove binder (MGB) and non fluorescent quencher (NFQ) on the 3' end.

    Hence, correct match is (a)-3, (b)-1, (c)-4, (d)-2.

144.    To investigate the dynamic nature of the two unrelated centrosome localized GFP-tagged proteins [GFP-A; GFP-B], a team of scientists conducted fluorescence recovery after photo bleaching (FRAP) experiment. The FRAP profile of these two proeins is given below :

    

    The following statements for this FRAP analysis were made

    P.    GFP-B shows faster exchnage rate than GFP-A.

    Q.    GFP-A shows faster exchange rate than GFP-B.

    R.    GFP-A has more immobile fraction than GFP-B.

    S.    GFP-B has more immobile fraction than GFP-A.

    Which of the above mentioned statements for GFP-A and GFP-B are correct ?

    (a)    P and R

    (b)    P and S

    (c)    Q and R

    (d)    Q and S

Ans.    (a)

Sol.    The FRAP analysis were made : GFP-B shows faster exchange rate than GFP-A and GFP-A has more immobile fraction than GFP-B.

    

145.    Area of patch 1 is 2000 m² with a resource density of 5 units/m². Area of patch 2 is 3000 m² with a resource density of 10 units/m². As per the theory of ideal-free distribution, organisms distribtue themselves such that the expected ratio of abndance of organisms in the two patches (patch 1 : patch 2) is

    (a)    1 : 2

    (b)    2 : 3

    (c)    1 : 3

    (d)    3 : 2

Ans.    (c)

Sol.    Given, area of patch 1 = 2000 m²

    Area of patch 2 = 3000 m²

    Resources density of patch 1 = 5 units/m²

    and Resources density of patch 2 = 10 units/m²

        For pathch 1 : 2000 × 5 = 10000

        For patch 2 : 3000 × 10 = 30000

        Ratio of patch 1 : patch 2 i.e.,