PYQ DEC 2017 - VedPrep

CSIR NET BIOLOGY (DEC - 2017)
Previous Year Question Paper with Solution.

21. Choose the correct statement from the following :

(a) Disulfide bonds in a 20-residue peptide can be formed only if the cysteines are adjacent to each other.

(b) The amino acid isoleucine has only one chiral centre.

(d) The pI of aspartic acid is less than that of glutamic acid.

Ans. (d)

Sol. The isoelectric point [pI, pH (I), IEP] is the pH at which a particular molecule carries no net electrical charge or electrically neutral in the statistical mean. Glutamic acid has one additional methylene group in its side chain than does aspartic acid. The side chain carboxyl of aspartic acid is referred to as the carboxyl group, while that of glutamic acid is referred to as the carboxyl group.

the pKa of the carboxyl group of glutamic acid in a polypeptide is about 4.3, significantly higher than that of aspartic acid. This is due to the inductive effect of the additional methylene group.

Table of pKa and pI values

The pKa values and the isoelectronic point, pI are given below for the 20 -amino acids.

pKa1 = -carboxyl group, pKa2 = -ammonium ion and pKa3 = side chain group.

22. Indicate the incorrect statement from the following :

(a) Allosteric enzymes function through reversible noncovalent binding of allosteric modulators or effectros.

(b) Monoclonal antibodies that catalyze hydrolysis of esters or carbonates can be produced.

(d) Acid phosphates hydrolyze biological phosphate esters at ~ pH 5.0.

Ans. (c)

Sol. Irreversible Inhibition of an Enzyme : Many enzymes are inhibited irreversibly by heavy-metal ions such as Hg²⁺, Cu²⁺, which can react with essential sulphydryl groups to form mercaptides :

Enz—SH + Ag⁺ Enz—S—Ag + H⁺

The affinity of Ag⁺ for sulphydryl groups is so great that Ag⁺ can be used to titrate—SH groups quantitatively.

23. The and values of a -strand composed of all D-amino acids will mainly occupy which quadrant in the Ramachandran plot?

(a) upper left

(b) upper right

(d) lower right

Ans. (d)

Sol. Ramachandran Plot is a way to visualize dihedral angles against of amino acid residues in protein structure.

Quadrant I shows a region where some conformations are allowed. This is where rare left-handed alpha helices lie.

Quadrant II shows the biggest region in the graph. This region has the most favourable conformations of atoms. It shows the sterically allowed conformations for beta strands.

Quadrant III shows the next biggest region in the graph. This is where right-handed alpha helices lie.

Quadrant IV has almost no outlined region. This conformation ( around –180 to 0 degrees, around 0-180 degrees) is disfavoured due to steric clash.

24. Match the coenzymes in column I serving as transient carriers of specific atoms or functional groups in column II

Select correct combinations from the options given below:

(a) A–(iv), B–(iii), C–(ii), D–(i)

(b) A–(iii), B–(iv), C–(i), D–(ii)

(d) A–(ii), B–(i), C–(iv), D–(iii)

Ans. (a)

Sol. Some Common Coenzyme

25. Out of the statements mentioned below

1. L-threonine and L-allo-threonine interact identically with plane polarized light.

2. van der Waals' interactions are always attractive.

3. Poly (pro) II-helix is not stabilized by intermolecular hydrogen bonds.

4. The folding of a protein is associated with an overall positive change in free energy and negative change in entropy.

5. Lysine acetylation on histone is associated with loosening of the histone complex from DNA.

Which of the following combinations is CORRECT?

(a) 1 and 3

(b) 2 and 4

(d) 4 and 5

Ans. (c)

Sol. Histone acetylation and deacetylation are the process by which the lysine residues within the N-terminal tail protrouding from the histone core of the nucleosome are acetylated and deacetylated as part of gene regulation. Lysine acetylation on histone is associated with loosening of the histone complex from DNA. A polyproline helix is a type of proteins comprising repeating proline residues. A left-handed polyproline II helix (PPII, poly-Pro II) is formed when sequential residues all adopt backbone dihedral angles of roughly (–75°, 150°) and have trans isomers of their peptide bonds. This PPII conformation is also common in proteins and polypeptides with other mino acids apart from proline. The PPII helix is relatively open and ahs no internal hydrogen bonding, as opposed to the more common helical secondary structures, the alpha helix and its relatives the 310 helix, as well as the -helix. It is not stabilized by intermolecular hydrogen bonds.

26. The following statements are made:

1. and anomers of glucose are interconvertible and the ratio of their abundance is 1:2, respectively.

2. Single chain lipids (≥C₁₄) form micelles and double chain lipids form bilayers in water.

3. Proline is energetically favoured at the Ctermini of an -helix than at the N-termini.

4. Major groove of DNA readily accommodates several common structural motifs in protein than the minor groove.

5. Replacement of a canonical Watson-Crick pairing by Wobble base pairs does not change the surface properties in t-RNA.

Which one of the following combinations is INCORRECT?

(a) 1 and 4

(b) 2 and 5

(d) 2 and 3

Ans. (c)

Sol. The abundance of alpha and beta anomers is not in the 1:2 ratio, but they are present as 64% and 36% respectively. It is however true that single chain lipids will form micelles and double chain lipids form bilayer.

27. From the following statements

P. Enzymes enhance reaction rates by a factor of 2 to 10.

Q. The activation energy for a reaction is lowered by enzymes.

R. The interactions between enzymes and substrates are hydrogen bonding, hydrophobic and ionic.

S. Substrate concentration does not affect the rate of enzyme catalyzed reactions.

Pick the combination with all INCORRECT statements.

1. P and Q

2. Q and R

3. P and R

4. P and S

Ans. (d)

Sol. Enzymes (and other catalysts) act by reducing the activation energy, thereby increasing the rate of reaction. Enzyme concentration does not affect the enzyme activity directly as substrate concentration determine it until the presence of the substrate in an excess amount. Enzymes are the assemblage of protein subunits and hence function under specific temperature and pH range only.

28. A polymer is synthesized from an achiral amino acid. Conformation of the polymer can be investigated by the following techniques.

P. Fibre diffraction

Q. Nuclear magnetic resonance spectroscopy

R. Circular dichroism spectroscopy

S. Differential scanning calorimetry

Choose the combination which would indicate that the polymer adopts a helical conformation

1. P and R

2. Q and S

3. P and Q

4. R and S

Ans. (c)

Sol. A conformational isoform of the mammalian prion protein (PrPSc) is the role compnent of the infectious pathogen that causes the prion diseases. We have obtained X-ray fiber diffraction patterns from infectious prions that show cross- diffraction : meridional intensity at 4.8 Å resolution, indicating the presence of strands running approximately at right angles to the filament axis and characteristic of amyloid structure. Some of the patterns also indicated the presence of a repeating unit along the fiber axis, corresponding to four -strands.

high resolution NMR can be used for polymer characteriation both in solution and in the solid state. The power of NMR as a method arises from the observation of individual carbon and proton atoms along the main chain and side chains of polymers. This makes it possible to characterize in detail it possible to characterize in detail the microstructure, the conformation and the molecular organization in the solid.

29. Following table lists the two major forms of DNA duplexes, conformation of base attached to the sugar and the nature of major and minor grooves.

Which of the following combination correctly depicts the types of DNA duplexes and their properties (numbered within bracket)?

(a) A – (i) (iii) (vi); B – (ii) (iii) (vi)

(b) A – (i) (iii) (vi); B – (ii) (iii) (v)

(d) A – (ii) (iv) (v); B – (i) (iii) (vi)

Ans. (c)

Sol. Three major forms of DNA are double stranded and connected by interactions between complementary base pairs. These are terms A-form, B-form and Z-form DNA. The major difference between A-form and B-form nucleic acid is in the conformation of the deoxyribose sugar ring.

Comparison of the Structural Properties of A-, B- and Z-DNA

30. Which one of the following statement about chromatin is NOT true?

(a) DNA winds approximately 1.65 timse around the nucleosomes

(b) H2A-H2B bind to both the entry and exit ends of DNA in nucleosomes

(d) Non-histone proteins are part of mitotic chromosomes

Ans. (b)

Sol. The convex surface of the H2A/H2B and H3/H4 heterodimers carries a strong positive charge and constitutes the primary DNA binding element of each histone-fold heterodimer. The histone octamer forms a spool for wrapping nucleosomal DNA. It is assmebled from two H3/H4 and two H2A/H2B histone-fold heterodimers using a single structural motif, the four helix bundle. Each four helix bundle is formed by the and helices from the adjacent histone folds.

31. During eukaryoutic cell division, metaphase to anaphase transition is regulated by degradation of

(a) Cyclin B1

(b) CDK1

(d) Polo-like kinase

Ans. (a)

Sol. The degradation of the cyclin B subunit of protein kinase Cdk1/cyclin B is required for inactivation of the kinase and exit from mitosis. Cyclin B is degraded by the ubiquitin pathway, a system involved in most selective protein degradation in eukaryotic cells.

In this pathway, proteins are targeted for degradation by ligation to ubiquitin, a process carried out by the sequential action of three enzymes : the ubiquitin-activating enzyme E1, a ubiquitin-carrier protein E2 and a ubiquitin-protein ligase E3. In the system responsible for cyclin B degradation, the E3-like function is carried out by a large complex called cyclosome or anaphase promoting complex (APC).

32. Sphingolipids (S) and cholesterol (C) molecules of the lipid bilayer aggregate into multiple tiny rafts instead of a single large one. Considering that size of a lipid raft depends on the affinity of S and C for one another and other lipids in the membrane, choose the option that best describes this property.

(a) S and C bind to one another tightly and independent of any other lipid molecules.

(b) S and C bind to one another with same affinity as they bind to other lipid species of the membrane

(d) S and C have slightly higher affinity than other lipid molecules of the membrane and are in dynamic equilibrium with their free forms.

Ans. (d)

Sol. Membrane microdomain enriched in glycosphingolipids and cholesterol and containing glycosylphosphatidylinositol (GPI) anchored proteins hvae been proposed as lateral structural components of the plasma membrane. These microdomain have been implicated in the polarized sorting of proteins and cellular signaling. A complementary view of sphingolipid-cholesterol rafts is that they form separate liquid-ordered phases in the bilayer, which are dispersed in the liquid-disordered matrix formed by unsaturated glycerophospholipids. The liquid-ordered phase in a lipid bilayer, as characterized by highly ordered carbon chains coupled with a high degree of rotational freedom.

33. To understand the microtubule-dependent motor activity of a freshly purified motor protein from budding yeast, a researcher set up microtubule based gliding assay. In such an assay where microtubules are fluorescently tagged at its (+) end, the researcher observed that this motor protein moves the microtubule in the direction of its (+) end as shown below:

The newly identified motor protein is

(a) Dynein

(b) Myosin

(d) Either Dynein or Kinesin 1

Ans. (c)

Sol. Dynein is a family of cytoskeletal motor proteins that move along microtubules in cells. They convert the chemical energy stored in ATP to mechanical work. Dynein transports various cellular cargos, provides forces and displacements important in mitosis and drives the beat of eukaryotic cilia and flagella.

34. Given below are organelles (column A) and properties associated with the organelles (column B).

Choose the option that matches the organelles with the most appropriate property.

(a) A – (iv) ; B – (iii); C – (v); D – (i); E – (ii)

(b) A – (v) ; B – (iv); C – (i); D – (v); E – (ii)

(d) A – (iv) ; B – (v); C – (ii); D – (i); E – (iii)

Ans. (a)

Sol. 1. The mannose 6-phosphate receptors (MPRs) are transmembrane glycoproteins that target enzymes to lysosomes in vertebrates. Mannose 6-phosphate receptors bind newly synthesized lysosomal hydrolases in the trans-Golgi network (TGN) and deliver them to pre-lysosomal compartments.

2. COPI is a coatomer, a protein complex that coats vesicles transporting proteins from the cis end of the Golgi complex back to the rough endoplasmic reticulum (ER), where they were originally synthesized and between Golgi compartments.

3. Secretory vesicles form from the trans Golgi network and they release their contents to the cell exterior by exocytosis in response to extracellular signals. The secreted product can be either a small molecule (such as histamine) or a protein (such as a hormone or digestive enzyme).

4. COPII is a coatomer, a type of vesicle coat protein that transports proteins from the rough endoplasmic reticulum to the Golgi apparatus. This process is termed anterograde transport, in contrast to the retrograde transport associated with the COPI protein. COPII proteins are required for selective export of newly synthesized proteins from the endoplasmic reticulum (ER).

5. The major route for endocytosis is most cells and the best-understood, is that mediated by the molecule clathrin. This large protein assists in the formation of a coated pit on the inner surface of the plasma membrane of the cell. This pit then buds into the cell to form a coated vesicle in the cytoplasm of the cell.

35. Cell cycle checkpoints are surveillance mechanisms that ensure order and fidelity of events of the cell cycle. Given below are some of the checkpoint proteins and their functions.

Match the checkpoint protein with its function.

(a) A – iv, B – i and ii, C – iii

(b) A – ii and iv, B – i, C – iii

(d) A – ii and iv, B – iii, C – i

Ans.

Sol. The normal function of MAD2 is to accumulate at kinetochores and generate a wait signal preventing the cell from progressing to anaphase of the cell cycle until the spindle microtubules have correctly aligned with the kinetochores on each chromosome. Mad2 is a pivotal component of the spindle assembly checkpoint (SAC) which inhibits anaphase promoting complex/cyclo-some (APC/C) activity by sequestering Cdc20 thereby regulating the destruction of securin and cyclin B.

ATM and ATR respond to different types of DNA lesions, to which they are recruited through specific co-factors; ATM responds primarily to DNA double-strand breaks (DSBs), whereas ATR protects the integrity of replicating chromosomes.

Tem1 prevent activation of cdc14.

36. To assess the impact of a newly identified drug when added to a culture of subconfluent HeLa cells, a researcher analyzes the fluorescence activated cell sorting (FACS) profile of untreated (- Drug) versus treated (+ Drug) cells.

Based on the FACS profile shown above, this drug inhibits

(a) G₁ phase of the cell cycle

(b) S phase of the cell cycle

(d) G₀ phase of cell cycle

Ans. (c)

Sol. Cells in late S and G2 phases were found to be maximally sensitive to inhibition of cell growth and colony formation after short exposure of drug.

37. You have discovered a new transposon, TnX, and would like to identify its mode of replication. A heteroduplex of the TnX sequence is made with a few mismatches and introduced into bacteria. The newly transposed genomic loci are sequenced. You find that the sequence of the transposon matched exactly with one of its parent strands. This suggests that

(a) TnX transposes by conservative transposition mechanism.

(b) TnX transposes using a site-specific recombination mechanism.

(d) TnX transposes by replicative mechanism.

Ans. (d)

Sol. If the starting DNA is heteroduplex, then the resulting DNA will still the heteroduplex only in a conservative or nonreplicative pathway. However, in a replicative pathway, transposition results in individual cells that are either of the one strand i.e. results in homoduplexes. Or if replication takes place (the replicative mode of transposition), all colonies are like either of the strand, because the replication will convert the heteroduplex DNA into two homoduplex daughter molecules.

38. Select a cellular body which is NOT a part of the nucelar bodies :

(a) P-bodies

(b) Nucleolus

(d) Interchromatin granule clusters

Ans. (a)

Sol. The cell nucleus is a complex organelle whose dynamic architecture consists of numerous subcellular compartments, collectively referred to as nuclear bodies. These structures include nucleoli, Cajal bodies (CBs), histone locus bodies (HLBs), splicing factor compartments (a.k.a. speckles or interchromatin granule clusters), paraspeckles, promyelocytic leukemia (PML) bodies, Gemini bodies (gems), perinucleolar compartments (PNCs), polycomb group (PcG) bodies, heat shock factor 1 (HSF1) foci, SAM-68gv5y4btg2432zd3dx bodies, GATA-1 foci and many more.

One of the emergent principle of nuclear organization is that certain subnuclear domains are associated with specific gene loci. Another important rule is that associations between these suubdomains and loci are dynamic and can change in response to cellular signals.

39. One of the mechanisms used by bacteria for adaptation to changed environment is altering transcription of their genes. In this regard, which one of the following responses is NOT found in bacteria?

(a) A gene with two different promoters for expression in different conditions

(b) Use of different sigma factors for transcription of genes

(d) Expression of anti-sigma factors

Ans. (c)

Sol. Altered transcription as per changing environment if achieved by sigma factors.

40. Which one of the following proteins acts both as an activator and repressor of transcription?

(a) cI protein

(b) N protein

(d) Q protein

Ans. (a)

Sol. The cI protein of bacteriophage lambda (lamda cI) is both a represor and activator of transcription that has served as a model for understanding how gene regulatory proteins work. A dimeric DNA-binding protein lambda cI also forms higher-order oligomers that allow it to bind cooperatively to both adjacent and nonadjacent operator sites within the phage genome. The ability of phage lambda to transition efficiently from one program of gene expression to another depends upon the formation of these higher order proein DNA complexes.

41. A highly specific inhibitor that targets the phosphorylation activity of TFIIH is added to an in vitro tarnscription reaction. Which one of the following steps is most liketly to be affected?

(a) Binding of RNA polymerase to promoter sequence

(b) Promotoer clearance

(d) Open promoter complex formation

Ans. (b)

Sol. FIR bnds with the central DNA-binding domain of FBP and interacts with the TFIIH complex, in vivo and in vitro. TFIIH complex, in vivo and in vitro. TFIIH is a multi-functional, multisubunit RNA polymerase II transcription factor required at several steps during transcription initiation and promoter clearance.

42. During eukaryotic protein synthesis, stress conditions result in activations of specific kinases leading to phosphorylation of a key translation initiation factor that inhibits protein synthesis from a large number of cellular mRNA. Which one of the following factors is the target of the kinase?

(a) eIF4E

(b) eIF4G

(c)

(d) Gcn4

Ans. (c)

Sol. A central mechanisms for translocational control involves phosphorylation of the subunit of eukaryotic initiation factor (eIF) 2 (eIF2-P), which represses the initiation phase of protein synthesis, allowing cells to conserve resources while a new gene expression program is adopted to prevent stress damage.

43. Imagine the following RNA sequence is translated in the mammalian cytosol and mitochondria.

RNA sequence:

AUG AUA CUG UGA CUU AGG CUC UAA

Following are some putative peptide sequences

Find out the correct combination of peptides made in the cytosol and mitochondria.

(a) (A) (i)

(b) (A) (ii)

(d) (B) (ii)

Ans. (b)

Sol. Mammalian Cytosol : AUG – Methoinine; AUA – isoleucine; CUG – Leucine; UGA – stop; Mitochondria; AUG – Methoinine; AUA – Methonine; CUG – Leucine; UGA – Trptophan; CUU – Leucine; AGG – Stop codon.

44. In all organisms, it is critical that replication initiation be tightly controlled to ensure that chromosome number and cell number remain appropriately balanced. Given below are several statements regarding regulation of replication in E. coli.

P. Hemimethylation and sequestration of oriC (origin of replication) by a protein called SeqA prevents initiation of replication.

Q. Availability of DnaA protein is an important requirement for initiation of replication.

R. The ratio of ADP : ATP is important as high level of ADP is required for initiation of replication.

S. Recruitment of Hda protein by sliding clamp inhibits ATP hydrolysis required for initiation of replication.

Which of the above statements are NOT true?

(a) P and Q

(b) Q and C

(d) P and S

Ans. (c)

Sol. In E. coli, origin usage is prevented by a process called origin sequestration that depends on the binding of the SeqA protein to newly replicated origins.

SeqA discriminates between new and old origins by the methylation status of GATC sequences, which are present at high frequency in the oriC DNA sequence. The adenines of these sequences are recognized and methylated by Dam methylase. During this time, the orgin is bound by SeqA, which in addition to preventing GATC methylation by Dam methylase, inhibits replication initiation. Hda, mediates the interaction of DnaA with the a^{^}-subunit sliding clamp to promote hydrolysis of DnaA-bound ATP as well as the cellular fraction containing IdaB. Hda is essential for the control of initiation of DNA replication by inhibiting reinitiation of replication.

45. A construct (shown below) was generated to access the activity of bidirectional promoter (A and B) which has a common regulatory DNA element (X). The construct was used to transform E. coli.

The activity of the promoter (as shown in the graph below) is recorded in the presence of increasing levels of an E. coli encoded regulatory protein which binds to DNA element X.

The above experiment is repeated in a mutant E. coli with mutation which abolishes binding of the regulatory protein to element X. Which one of the following graphs best depicts the activity of the reporter genes in the mutant strain?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Option (d) represents correct graph for both reporter genes.

46. In E. coli, though DNA polymerase I (Pol I) plays an essential role in the replication process, it is not the major polymerase. Instead, the enzyme responsible for advancement of replication fork is Pol III. From the four DNA structures (A, B, C and D) given below, students made several interpretations about the shorter arm being extended by Pol I and/or Pol III.

Which one of the interpretation written below is correct?

(a) P will be extended by Pol III but not by Pol I.

(b) Neither Q nor R will be extended by either Pol I or Pol III.

(d) S will be extended only by Pol I, but not by Pol III.

Ans. (b)

Sol. Option Q and R can't be extended by any DNA polymerase because their 3' OH end already exceeds in size than template strand.

47. tRNA genes of a Gram positive bacteria were sequenced. While some of the genes possessed sequence corresponding to the –CCA end of the tRNA, others did not. Interestingly, even the genes that lacked the sequence corresponding to the –CCA end of the tRNA were found to code for a functional tRNA. This is because

(a) the –CCA end in the tRNA is not essential for its function.

(b) the genes that do not possess sequence corresponding to the –CCA ends of the tRNA are repaired prior to their transcription.

(d) the primary transcripts of the genes lacking sequence corresponding to the – CCA end are subjected to g-RNA mediated editing prior to their maturation.

Ans. (c)

Sol. Eukaryotic tRNA transcripts lack the obligatory CCA sequence at their 3' ends. This is appended to the immature tRNAs by the enzyme CCA-adding polymerase, which sequentially adds two C's and an A to tRNA using CTP and ATP as substrates post transcriptionally.

48. In a signalling event, binding of an extracellual ligand activates G-protein coupled receptor (GPCR) that eventually activates phospholipase C-. Which one of the following statements truly reflects the function of phospholipase C- ?

(a) Phospholipase C- converts PI (3, 4, 5)P₃ to PI (4, 5)P₂

(b) Phospholipase C- converts PI (4)P to PI (4, 5)P₂

(d) Phospholipase C- converts PI (5)P to PI (4, 5)P₂

Ans. (c)

Sol. Transient receptor potential classical (or canonical) (TRPC)3, TRPC6 and TRPC7 are a subfamily of TRPC channels activated by diacylglycerol (DAG) produced through the hydrolysis of phosphatidylinositol 4, 5-bisphosphate [PI(4,5)P2] by phospholipase C(PLC). PI(4,5)P2 depletion by a heterologously expressed phosphatase inhibits TRPC3, TRPC6 and TRPC7 activity independently of DAG; however, the physiological role of PI(4,5)P2 reduction on channel activity remains unclear.

49. Proteins with cytoplasmic domains having tyrosine kinase activity do NOT act as receptors for

(a) Epidermal growth factor (EGF)

(b) Platelet-derived growth factor (PDGF)

(d) Transferrin

Ans. (d)

Sol. The transferrin receptor is a major protein found on the basolateral membranes of intestinal epithelial cells, yet its possible role in intestinal iron metabolism and also in iron absorption unclear.

50. Junctions which tether cytoskeletal filaments inside the cell are known as

(a) anchorgine junctions

(b) occluding junctions

(d) signal–relaying junctions

Ans. (a)

Sol. Anchoring junctions are cell junctions that are anchored to one another and attached to components of the extracellular matrix. They are important in keeping the cells togethera nd structural cohesion of tissues. They are commonly found in tissue that are prone to constant mechanical stress, e.g., skin and heart.

There are three types of anchoring junctions. They are desmosomes, hemidesmosomes and adherens junctions. Both desmosomes and hemidesmosomes use intermediate filaments as their cytoskeletal anchor. However, the transmembrane linker of desmosomesis cadherin whereas that of hemidesmosomes is integrins. Desmosomes link cell to anothe cell.

Hemidesmosomes link cell to the extracellular matrix. The adherens junctions use actin filaments as their cytoskeletal anchor. Their transmembrane linker is cadherin or integrin.

51. Which one of the following human serum immunoglobulins takes part in classical complement fixation pathway?

(a) IgD

(b) IgE

(d) IgG

Ans. (d)

Sol. The clasical complement pathway is one of three pathways which activate the complement system, which is part of the immune system. The classical complement pathway is initiated by antigen-antibody complexes with the antibody complexes with the antibody isotypes IgG and IgM.

52. In a type of signal transduction pathway, ligand binding to a receptor triggers activation of a receptor-associated kinase. This kinase may be an intrinsic part of the receptor protein or tightly bound to the receptor. Receptors in which the tyrosine kinase is an intrinsic part of it's polypeptide chain are called the receptor tyrosine kinase (RTK). Which one of the following statements regarding RTK is INCORRECT?

(a) All RTKs have three essential components: an extracellular domain containing ligand binding site, a transmembrane domain and a cytoplasmic segment that includes a domain with protein tyrosine kinase activity.

(b) Most RTKs are monomeric and ligand binding to the extracellular domain induces formation of receptor dimers.

(d) Ligand binding to RTK leads to autophosphorylation of the protein tyrosine kinase in the cytoplasmic domain. The activated kinase then phosphorylates several tyrosine residues in the receptor's cytoplasmic domain.

Ans. (c)

Sol. Receptor tyrosine kinases (RTKs) are a subclass of tyrosine kinases that are involved in mediating cell-to-cell communication and controlling a wide range of complex biological functions, including cell growth, motility, differentiation, and metabolism. The RTK family includes, among others, epidermal growth factor receptor (EGFR), platelet-derived growth factor receptors, fibroblast growth factor receptors (FGFRs), vascular endothelial growth factor receptors.

53. Two classes of genes - proto-oncogene and tumor suppressor gene usually contribute to the development of cancer. Following are some of the statements regarding both the genes.

A. Proto-oncogenes result in the development of cancer by gain-offunction mutation whereas tumor suppressor gene leads to cancer development by loss-of-function mutation.

B. Proto-oncogenes result in development of cancer by loss-of-function mutation whereas tumor suppressor gene leads to cancer development by gain-of-function mutation.

C. Mutation in both the alleles of a protooncogene is required for induction of cancer whereas mutation in one of the two alleles in tumor suppressor gene is sufficient for promoting tumorigenesis.

D. Mutation in one of the two alleles in proto-oncogene is sufficient for induction of cancer whereas mutation in both the alleles of a tumor suppressor gene is required for promoting tumorigenesis.

Which combinations of the above statements are true for both the genes?

(a) A and B

(b) A and C

(d) B and C

Ans. (c)

Sol. Two broad classes of genes-protooncogenes (e.g., ras) and tumor-suppressor genes (e.g., APC)—play a key role in cancer induction. These genes encode many kinds of proteins that help control cell growth and proliferation; mutations in these genes can contribute to the development of cancer.

Most cancers have inactivating mutations in one or more proteins that normally function to restrict progression through the G1 stage of the cell cycle.

Conversion or activation of a protooncogene into an oncogene generally involves a gain-of-function mutation. At least three mechanisms can produce oncogenes from the corresponding proto-oncogenes.

Point mutations in a proto-oncogene that result in a constitutively acting protein product.

Localized reduplication (gene amplification) of a DNA segment that includes a proto-oncogene, leading to overexpression of the encoded protein.

Chromosomal translocation that brings a growth-regulatory gene under the control of a different promoter and that causes inappropriate expression of the gene.

Tumor suppressor genes generally encode proteins that in one way or another inhibit cell proliferation.

Loss of one or more of these "brakes" contributes to the development of many cancers. Five broad classes of proteins are generally recognized as being encoded by tumor-suppressor genes :

Intracellular proteins, such as the p16 cyclin-kinase inhibitor, that regulate or inhibit progression through a specific stage of the cell cycle.

Receptors for secreted hormones (e.g., tumor derived growth factor ) that function to inhibit cell proliferation.

Checkpoint-control proteins that arrest the cell cycle if DNA is damaged or chromosomes are abnormal.

Proteins that promote apoptosis.

Enzymes that participate in DNA repair.

54. Two important features which aid the development of a tumor and its metastasis are epithelial-to-mesenchymal transition and angiogenesis. A student tested four cell lines to determine their invasiveness and proliferation capability by checking the expression of VEGF-A, TWIST and Cyclin D1. Which one of the following figures is most likely to exhibit the characteristics of a highly metastatic cancer cell?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Vascular endothelial growth factor (VEGF) is a proangiogenic factor that is up-regulated in various tumors. VEGF has been shown to interact with COX-derived prostaglandins in angiogenesis. Cyclin D1 gene overexpression and amplification have been shown to play a role as prognostic factors in many human cancers.

55. Several types of molecules including the transmembrane glycoproteins can function as matrix receptors and co-receptors. However, the principal receptors on animal cells for binding most extracellular matrix proteins are the integrins. Which of the following statements is NOT true for integrins?

(a) Integrins are transmembrane linker proteins that link to the cytoskeleton.

(b) An integrin molelcule is composed of two non-covalently associated glycoprotein subunits and . Both subunits span the cell membrane, with short intracellular C-terminal tails and large N-terminal extracellular domains.

(c) The extracellular portion of the integrin dimer binds to specific carbohydrate residues in extracellular matrix proteins or to ligands on the surface of other cells.

(d) The intracellular portion binds to a complex of proteins that form a linkage to the cytoskeleton.

Ans. (c)

Sol. Integrins are large, membrane-spanning, heterodimeric proteins that are essential for a metazoan existence. All members of the integrin family adopt a shape that resembles a large "head" on two "legs", with the head containing the sites for ligand binding and subunit association. Most of the receptor dimer is extracellular, but both subunits traverse the plasma membrane and terminate in short cytoplasmic domains.

These domains initiate the assembly of large signaling complexes and thereby bridge the extracellular matrix to the intracellular cytoskeleton. Integrins link the actin cytoskeleton of a cell to various external structures.

The cytoplasmic portion of each integrin molecules bind to the actin filaments inside the cell. The extracellular portion of the integrin then binds to molecules in the extracellular matrix or on the surface of other cells. Integrin attachments to neighbouring cells can break and reform as a cell moves.

56. Major histocompatibility complex (MHC) molecules are encoded by a cluster of genes called MHC locus. There are several reasons why an MHC molecule on the surface of a cell is important. Which one of the following reasons is INCORRECT?

(a) To display self class I to demonstrate that the cell is normal and healthy.

(b) To display foreign-peptide in class I to show that the cell is infected and to engage with T helper cells.

(d) To display a self-peptide in class I and II to maintain tolerance to self-proteins.

Ans. (b)

Sol. The function of MHC molecules is to bind peptide fragments derived from pathogens and display them on the cell surface for recognition by the appropriate T cells.

57. Oncogenic viruses could have either DNA or RNA genomes. Listed below are some oncogenic viruses (Column A), their genome types (Column B) and the cancers caused by these viruses (Column C).

Find out the correct combination.

(a) P-1, X; Q-2, Y; R-2, Z

(b) P-2, Y; Q-1, Z; R-2, X

(d) P-1, Z; Q-1, X; R-2, Y

Ans. (d)

Sol. Chronic infection with the hepatitis B virus has been linked epidemiologically to the developmetn of hepatocellular carcinoma for more than 30 years.

Burkitt lymphoma (BL) is an aggressive B-cell malignancy with endemic, sporadic and immunodeficiency-associated variants. Endemic EBV associated BL has an incidence of 5-10/100,000 children and accounts for up to 74% of childhood malignancies in the African equatorial belt.

Human T-cell leukemia virus type I (HTLV-1) and HTLV-2 are complex retro-viruses that persist in the host, eventually causing leukemia and neurological disease in a small percentage of infected individuals. In addition to structural and enzymatic proteins, HTLV encodes regulatory (Tax and Rex) and accessory.

58. Which one of the following is a group of signalling molecules that act as morphogens during development of an organism and its effects are mediated through the receptor Patched and its binding partner Smoothened?

(a) Hedgehog protein

(b) Notch protein

(d) -catenin

Ans. (a)

Sol. During development, cells need to differentiate and perform different functions in order to permit the eventual formation of a mature organism. Receptors on cell surfaces play a leading role in mediating the ability of celsl to respond to activating signals during the development of an organism. Hedgehog (Hh) represents a family of signaling molecules that control decisions regarding development.

59. Match the following cleavage patterns with the species in which they occur.

(a) A–iv, B–iii, C–i, D–ii

(b) A–iii, B–i, C–iv, D–ii

(d) A–ii, B–iv, C–iii, D–i

Ans. (a)

Sol. (a) Spiral holoblastic cleavage is characteristic of several animal groups, including annelid worms, some flatworms and most molluscs. It differs from radial cleavage in numerous ways. First, the cleavage planes are not parallel or perpendicular to the animal-vegetal axis of the egg; rather, cleavage is at oblique angles, forming a "spiral" arrangement of daughter blastomeres. Second the cells touch one another at more places than do those of readially cleaving embryos.

(b) Holoblastic and radially symmetrical cleavage takes place in forg and salamander embryos, just similar to echinoderm cleavage. In the amphibian egg, yolk is concentrated in the vegetal hemisphere, whic is an barrier to cleavage. As a result first division starts at the animal pole and gradually takes place into the vegetal region.

Meroblastic : In the presence of a large amount of yolk in the fertilized egg cell, the cell can undergo partial or meroblastic, cleavage. Two major types of meroblastic cleavage are discodial and superficial.

(c) Discoidal : In discoidal cleavage, the cleavage furrows do not penetrate the yolk. The embryo forms a disc of cells, calleda blastodisc, on top of the yolk. Discoidal cleavage is commonly found in monotremes, birds, reptiles and fish that have telolecithal egg cells (egg cells with the yolk concentrated at one end). The layer of cells that have incompletely divided and are in contact with the yolk are called the "syncytial layer".

(d) Superficial : In superficial cleavage, mitosis occurs but not cytokinesis, resulting in a polynuclear cell. With the yolk positioned in the center of the egg cell, the nuclei migrate to the periphery of the egg and the plasma membrane grows inward, partitioning the nuclei into individual cells. Superficial cleavage occurs in arthropods that have centrolecithal egg cells (egg cells with the yolk located in the center of the cell). This type of cleavage can work to promote synchronicity in developmental timing, such as in Drosophila.

60. Which one of the followign statements regarding limb development in mice is true?

(a) The gene encoding Tbx5 is transcribed in the limb fields of the hindlimbs.

(b) The gene encoding Tbx4 is transcribed in the limb fields of the forelimbs.

(d) Genes encoding Islet 1, Tbx4 are Pitx are expressed in the presumptive forelimb.

Ans. (c)

Sol. Tbx4 gives identity to the limb bud of the hindlimb. Its forced expression can convert forelimb bud into hindlimb.

61. Based on ABC model during flower development, loss of class A activity results in the formation of only stamen and carpel. Which of the following floral organ identity genes controls the class A activity?

(a) APETALA 1 and APETALA 2

(b) APETALA 3 an PISTILLATA

(d) Only AGAMOUS

Ans. (a)

Sol. Class A genes, when expressed alone, produce sepals. The expression of classes A and B together directs petal identity. APETALA 1 and APETALA 2 control class A activity.

62. In a given experiment, transplantation of micromeres from the vegetal pole of a 16-cell sea urchin embryo onto the animal pole of a host 16-cell sea urcin embryo would initiate :

(a) the transplanted micromeres to invaginate into the blastocoel to create a new set of sekletogenic mesencyme cells.

(b) the transplanted micrometers to ingress into the blastocoel to create a new set of skeletogenic mesenchymal cells.

(c) the transplanted micromeres will mingle with the host micromeres to ingress into the blastocoel to create skeletogenic mesenchyme cells.

(d) the transplanted micromeres will form the secondary archenteron.

Ans. (a)

Sol. The micromeres undergo autonomous specification to become skeletogenic mesenchyme, and these micromeres produce the initial signals that specify the other tiers of cells. The micromeres are a major signaling center of the sea urchin embryo and their removal either at the 16-cell stage, or in blastulae, results in significant compensatory development and fate transitions within the embryo.

63. Which one of the following is NOT a phenotype of dark-grown seedlings that are etiolated?

(a) Short hypocotyls

(b) An apical hook

(d) Non-phosynthesis proplastids

Ans. (a)

Sol. Etiolation symptoms include absence of greening (Lack of chloroplast), reduced leaf size, hypocotyl elongation and maintenance of the apical hook.

64. During vulva development in C. elegans, the anchor cell produces Lin-3 protein which interacts with the Let-23 protein present on the six vulval precursor cells (VPCs) that form an equivalence group. The central lineage cell (P6.p) adopts the primary fate, the adjacent VPCs (P5.p and P7.p) adopt the secondary fate and the rest VPCs adopt the tertiary fate. Few mutants (Column A) and phenotypes (Column B) are listed in the table given below.

Match the correct mutant with the observed phenotype.

(a) A–iv, B–ii, C–iii, D–i

(b) A–iv, B–iii, C–i, D–ii

(d) A–iii, B–i, C–ii, D–iv

Ans. (a)

Sol. "All VPC which do not receive the anchor cell signal turn into hypodermal cells-tertiary fate-A-(iv) Lin3 if overexpresed, P5, 6, 7 adopt primary and P4 and P8 adopt secondary function-D-(i) Reduced lin-3 function creates one primary fate out of P6 and rest all are tertiary fate-C-(iii)".

65. C. elegans embryo uses both autonomous and conditional modes of specification. Conditional specification at the 4-cell stage can be seen in the development of the endoderm cell lineage and also in the establishment of dorsal-ventral axis. Following are few statements regarding this:

A. If the P₂ cell is removed at the early 4-cell stage, the EMS cell will divide into two MS cells and no endoderm will be made.

B. In pop-1 deficient embryos, both EMS daughter cells become E cells.

C. When the position of ABa and ABp was reversed, their fates get reversed and no normal embryo forms.

D. In embryos whose mother have mutant glp-1, ABp is transformed into ABa cell.

Which of the above statements are true?

(a) A, B and D

(b) A, B and C

(d) A, C and D

Ans. (a)

Sol. Both modes can be seen if the first two blastomeres are experimentally separated. The P1 cell develops autonomously without the presence of AB. It makes all the cells that it would normally make, and the result is the posterior half of an embryo. However, the AB cell, in isolation, makes only a fraction of the cell types that it would normally make. In pop-1-deficient embryos, both EMS daughter cells become E cells . In embryos whose mother have mutant glp-1 , ABp is transformed into ABa cell.

66. Following are the events that might take place during dorso-ventral axis specification in early embryonic development of Drosophila:

A. 'Torpedo' receptor activation

B. 'Pipe' synthesis

C. A cascade of protease activity

D. 'Cactus' dephosphorylation

E. Entry of 'Dorsal' in the nuclei of syncytial blastoderm stage embryo.

Which combination of the above events will occur in the presumptive dorsal side of the embryo deficient in maternal gurken?

(a) A only

(b) B and C only

(d) B, C, D and E only

Ans. (c)

Sol. Expression of 'Pipe' protein, cascade of protease activity and dephosphorylation of 'Cactus' are some of the major steps during dorsal-ventral axis specification in early embryonic development of Drosophila.

67. What would be the effect on newt limb regeneration, if more than 90% of the nerve supply is severed before amputation?

(a) The apical ectodermal cap stimulates growth of the blastema by secreting FGF8 but regeneration does not take place.

(b) Limb regeneration will take place and form a limb with no nerve supply.

(d) Limb regeneration with nerve supply will take place.

Ans. (a)

Sol. Regeneration is the process of renewal, restoration and growth that makes genomes, cells, organisms and ecosystems resilient to natural fluctuations or events that cause disturbance or damage.

Although the inhibitory effect of denervation on newt limb regeneration results in decreased macromolecular synthesis and changes in transcription, no specific protein has been identified to be differentially affected by denervation.

The effect on newt limb regeneration, if more than 90% of the nerve supply is severed before amputation is. The apical ectodermal cap stimulates growth of the blastema by secreting FGF8 but regeneration does not take place.

68. Fertilization in sea urchin involves interaction of sperm Bindin with its receptor EBR1, a 350 kDa glycoprotein on the egg vitelline membrane. The plot given below shows the status of membrane potential and levels of EBR1, Na+ and K+ in an unfertilized egg.

Which one of the following graphs best represents the condition within an egg 1-3 seconds after fertilization?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Sodium is essential for several physiologic processes related to fertilization and developmental in sea urchins, in addition to the previously described rapid-electrical block to polyspermy and the coupling of early and late events in egg activation. Sodium content changes with a biphasic time course due to the appearance of two exchange mechanisms:

1. A transient Na⁺/H⁺ exchange with a l/l stoichiometry causes a rapid rise in intracellular sodium. This exchange lasts about 3 min. and is not inhibited by poisoning the eggs with NaCN.

2. An Na⁺/K⁺ exchange, detectable within minutes of fertilization causes sodium content of the egg to fall and settle below unfertilized level.

69. What is the effect of sudden increase in the levels of ATP and citrate on an erythrocyte undergoing glycolysis?

(a) It inhibits glycolysis.

(b) It stimulates glycolysis.

(d) The rate of glycolysis increase gradually.

Ans. (a)

Sol. Phosphofructokinase-1 (PFK-1) is one of the most important regulatory enzymes of glycolysis. It is an allosteric enzyme made of 4 subunits and controlled by many activators and inhibitors. PFK is able to regulate glycolysis in response to the cell's energy requirements. For example, a high ratio of ATP to ADP will inhibit PFK and glycolysis.

PFK1 is allosterically inhibited by PEP, citrate and ATP. Phosphoenolypyruvic acid is a product further downstream the glycolytic pathway. Although citrate does build up when the Krebs Cycle enzymes approach their maximum velocity, it is questionable whether citrate accumulates to a sufficient concentration to inhibit PFK-1 under normal physiological conditions. ATP concentration build up indicates an excessof energy and does have an allosteric modulation site on PFK1 where it decreases the affinity of PFK1 for its substrate.

70. Which one of the following is the correct functions of JAZ (jasmonate zim-domain) protein family, a key regulator of Jasmonic Acid (JA) singnalling response?

(a) Binds of MYC2 and represses the JA dependent genes.

(b) Binds of MYC2 and transcribes JA dependent genes.

(d) Involved in directly inducing JA dependent genes.

Ans. (a)

Sol. JAZ proteins function to inhibit the activity of transcription factors responsible for driving the expression of JA-responsive target genes.

71. The plant hormone, Gibberellic Acid is generally NOT associated with

(a) stem elongation

(b) parthenocarpy

(d) malt production

Ans. (c)

Sol. Gibberllin is involved in stem elongation, malting of barley and promotion of seedless fruits (parthenocarpy). Parthenogenesis is a method of asexual reproduction, with embryo formation without fertilization and parthenocarpy involves fruit formation, without seed formation. The plant equivalent of parthenogenesis is apomixis.

72. Which one of the following statements is NOT correct for Casparian strip?

(a) It breaks the continuity of water movement through the apoplast pathway

(b) It is formed in the growing part of the root

(d) It is a band within the radial cell walls of the endodermis that is impregnated with lignin

Ans. (b)

Sol. The Casparian strip is a band of radial cell walls in the endodermis that is impregnated with the wax like, hydrophobic substance suberin. Suberin acts as a barrier to water and solute movement.

The endodermis becomes suberized in the non growing part of the root, several millimeters behind the root tip, at about the same time that the first protoxylem elements mature.

73. Which one of the following statements is true about xylem in plants?

(a) It is characterized by the presence of tracheary elements responsible for the conduction of water in vascular plants

(b) It is responsible for the transport of water and characterized by sieve elements like parenchyma cells and sclereids

(d) It is responsible for the transport of assimilates in both vascular and nonvascular plants

Ans. (a)

Sol. The term tracheary elements includes the two basic types of water-conducting cells in the xylem of vascular plants : tracheids and vessel elements. Tracheids differ from vessel elements; vessel elements have perforated end walls, whereas tracheids have primary wall material present on their end walls; both have lignified secondary walls and both can occur in primary and secondary xylem.

Tracheids are elongated cells with blunt ends, present along the long axis of the plant system. PHylogenetically the tracehids are most primitive type of cell found in xylem. These are imperforate or non-perforate cells having only pit-pairs on their common walls with lignified secondary wall. Tracheids are imperforate cells with bordered pits on their end walls.

The tracheids contains simple pits along with bordered pits. They are arranged one above the other and are present in both primary and secondary xylem with broader lumen that that of fibres. This broader lumen in tracheids and occurrence of the pits help in water conduction with dissolved mineral salts by acting as conduits in vascular plants. The presence of thick hard lignified wall offer mechanical support to the plants. Sometimes an intermediate type of cell element is also found in vascular system known as fibre-tracheids.

74. Match the chemical agents that interfere in oxidative phosphorylation process with their respective mode of action.

Choose the correct combination from below:

(a) A–(ii), B–(iv), C–(v), D–(iii)

(b) A–(iv), B–(i), C–(ii), D–(iii)

(d) A–(v), B–(ii), C–(i), D–(ii)

Ans. (b)

Sol. Antimycins are a group of secondary metabolites produced by Streptomyces bacteria. Antimycin A is the active ingredient in Fintrol, a chemical piscicide (fish poison) used in fisheries mangement. It blocks electron transfer from cytochrome b to cytochrome c1.

Oligomycins are macrolides created by Streptomyces that can be poisonous to other organisms. Oligomycin A inhibits ATP synthase by blocking its proton channel (F₀ subunit), which is necessary for oxidative phosphorylation of ADP to ATP (energy production).

The inhibition of ATP synthesis by oligomycin A will significantly reduce electron flow through the electron transport chain; however, electrons flow in not stopped completely due to a process known as proton leak or mitochondrial uncoupling.

Valinomycin is a naturally occurring dodecadepsipeptide used in the transport of potassium and as an antibiotic. It is a member of the group of natural neutral ionophorse because it does not have a residual charge. It Disrupts inner mitochondrial membrane potential.

Valinomycin was recently reported to be the must potent agent against severe acute respiratory syndrome coronavirus (SARS-CoV) in infected Vero E6 cells. Valinomycin acts as a nonmetallic isoforming agent in potassium selective electrodes.

Retenone is an odorless, colourless, crystalline isoflavone used as a broad-spectrum insecticide, piscicide and pesticide. Rotenon works by interfering with the electron transport chain in mitochondria. It inhibits the transfer of electrons from iron-sulphur centers in complex I to ubiquinone. This interferes with NADH during the creation of usable cellular energy (ATP).

Complex I is unable to pass off its electron to CoQ, creating a back-up of electrons within the mitochondrial matrix. Cellular oxygen is reduced to the radical, creating reactive oxygen species, which can damage DNA and other components of the mitochondria. Retenone also inhibits microtubule assembly.

75. A specific Mitogen Activated Protein Kinase (MAPK) cascade comprising MEKK1(a MAP3K), MKK1 (a MAP2K) and MPK6 (a MAPK) is activated sequentially in that order in Arabidopsis plants upon perceiving certain abiotic stress stimuli. The activated MPK6 phosphorylates and activates a transcription factor 'X', thereby making plant tolerant to the abiotic stress. Two different variants of MKK1 protein, a kinase inactive (KI) and a constitutively active (CA) forms were expressed independently in mkk1 mutant Arabidopsis plant. Considering above facts, which one of the following statements is CORRECT?

(a) 'X' will be activated even in the absence of stimuli in CA plants

(b) 'X' will not be activated in the absence of stimuli in CA plants

(d) The KI plants will be tolerant to the abiotic stress.

Ans. (a)

Sol. MKK1 and MKK2 function together with MPK4 and MEKK1 in a MAP kinase cascade to negatively regulate innate immune responses in plants. 'X' will be activated even in absence of stimuli in CA plants. Hu

76. Given below are some statements on secondary metabolites in plants.

A. Glucosinolates are synthesized by elongation in the length of side chains of their precursor amino acids.

B. All terpenes produced in herbs and spices are sticky, oily liquids which reduces the palatability (edible value) of these plants.

C. Cyanogenic glycosides are produced and stored in a toxic form in plants and are therefore more effective in defense against invading pathogens and herbivores.

D. The defense molecules, alkaloids (or their precursors), are gathered from plants and used by some insects for their own protection against predators.

Which one of the following represents correct statements?

(a) B and C only

(b) A and B only

(d) A and D only

Ans. (d)

Sol. Secondary metabolities are chemicals produced by plants for which no role has yet been found in growth, photosynthesis, reproduction or other "primary" functions.

These chemicals are extremely diverse; many thousands have been identified in several major classes. Each plant family, genus and sepcies produces a characteristic mix of these chemicals and they can sometimes be used as taxonomic characters in classifying plants. Humans use some of these compounds as medicines, flavourings, or recreational drugs.

In secondary metabolities Glucosinolates are synthesized by elongation in the length of side chains of their precursor amino acids. The defense molecules, alkaloids (or their precursors) are gathered from plants and used by some insects for their own protection against predators.

77. A researcher developed quadruple mutant that disrupted the function of all phytochrome interacting factor (PIF) family members. The following hypotheses were proposed regarding the phenotype of the mutant plants when grown in dark:

A. Plants would show short hypocotyls

B. Plants would be etiolated

C. Light induced genes would be activated

D. The cotyledons would be open

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

(a) A, B and C

(b) A, B and D

(d) B, C and D

Ans. (c)

Sol. The basic helix-loop-helix domain-containing transcription factors that interact physically with the red and far-red light photoreceptors, phytochromes, are called PHYTOCHROME INTERACTING FACTORS (PIFs). PIFs function as negative regulators of light responses by repressing photomorphogenesis and maintaining the skotomorphogenic state of the etiolated seedlings in darkness. Upon exposure to light, phytochromes promote the turnover of PIFs through rapid phosphorylation, ubiquitination, and proteasome-mediated degradation, which inhibits PIF function to induce transcriptional reprogramming, resulting in photomorphogenic development.

78. The photon intensity captured by green plants is used in the following processes:

(i) Photosynthesis

(ii) Generation of heat

(iii) Production of toxic products such as superoxide, singlet oxygen etc.

(iv) Damage to D1 protein of PSII

Based on the above facts, photoinhibiton will happen when the

(a) entire photon intensity is used for photosynthesis

(b) excess photon intensity is completely used for heat generation

(d) excess photon intensity leads to the damage of D1 protein

Ans. (d)

Sol. Photoinhibiton is light-induced reduction in the photosynthetic capacity of a plant, alga or cyanobacterium. Photosystem II (PSII) is more sensitive to light than the rest of the photosynthetic machinery and most researchers define the term as light-induced damage to PSII.

In living organisms, photoinhibited PSII centres are continuously repaired via degradation and synthesis of the D1 protein of the photosynthetic reaction center of PSII. It happen when excess photon intensity leads to the damage of D1 protein.

The photosynthetic efficiency is the fraction of light energy converted into chemical energy during photosynthesis in plants and algae. Phososynthesis can be described by the simplified chemical reaction

6H₂O + 6CO₂ + energy C₆H₁₂O₆ + 6O₂

Photon intensity captured by green plants is used in the following processes :

1. Photosynthesis

2. Generation of heat

3. Production of toxic products such as superoxide, singlet oxgyen etc.

4. Damage to D1 protein of PSII

79. With reference to plant biotic interactions, match the terms of Column I with the most appropriate term of Column II

(a) A – (ii), B – (iv), C – (i), D – (iii)

(b) A – (iii), B – (i), C – (ii), D – (iv)

(d) A – (iv), B – (i), C – (ii), D – (iii)

Ans. (d)

Sol. Phytoliths are commonly found in plants in the leaf epidermis and outermost covering of seeds and fruits, the epidermis of bracts which surround and protect grass seeds, and in the subepidermal tissue of orchid and palm leaves. Provide mechanical barrier to herbivory. Phloem feeders are known to induce plant resistance via the salicylic acid pathway, whereas biting-chewing herbivores induce plant resistance mainly via the jasmonate pathway. Recent studies reveal that plants close stomata upon guard cell perception of molecular signatures from microbes, microbe associated molecular patterns (MAMPs), to prevent microbe invasion. The perception of MAMPs induces signal transduction including recruitment of second messengers, such as Ca²⁺ and H₂O₂, phosphorylation events, and change of transporter activity, leading to stomatal movement. ETI can be triggered by toxins that are either directly injected into the host by bacterial secretion systems or internalized from the extracellular environment by endocytosis.

80. Transgenic tobacco plants over-expressing isopentenyl transferase (IPT) under the control of promoter region of Senescence-Associated Receptor kinase (P_SARK) were exposed to drought for 15 days followed by re-watering for 7 days. The following hypotheses were proposed regarding changes in the transgenic plants at the end of 7 days of re-watering:

A. The plants would be wilted and fail to survive.

B. The plants would be healthy and survive.

C. The plants would show higher production of cytokinin compared to wild type plants.

D. The plants would show higher production of absicic acid compared to wild type plants.

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

(a) A and C

(b) A and D

(d) B and D

Ans. (c)

Sol. Drought stress limits root growth and inhibits cytokinin (CK) production. Isopentenyl transferase gene codes for enzyme that catalyses first step in cytokinin biosynthesis. Increase in CK production through overexpression of isopentenyltransferase (ipt) alleviate drought damages to promote root growth. pSARK promoter makes sure that this gene is activated only in senescence stage and under drought.

So, the transgenic tobacco plants over-expression IPT gene will have enhanced production of cytokinin under drought and will be healthy and survive under periods of drought stress. Transgenic plants recover from drought stress earlier than wild type plants.

81. Which one of the following is NOT involved with the absorption of iron in the intestine?

(a) Divalent metal transporter

(b) Ferroportin 1

(d) Transferrin

Ans. (d)

Sol. Absorption of iron is one of the first steps in iron metabolism. Metabolism is a process of chemical interactions that generate energy from food that you eat. Iron metabolism is the part of the process that manages iron in the body. Iron enters the stomach where it is exposed to stomach acid and changed into a form that allows it to be absorbed. The portion of the small intestine called the duodenum is the chief area where iron absorption takes place. There may be a second minor absorption site near the end of the small intestinal tract.

Transferrin is not invovled with the absorption of iron in the intestine. Once iron is absorbed it is carried (transported) by a protein called transferrin. Each molecule of transferrin can carry two atoms of iron. When working normally, transferrin binds to iron and transports it to all tissues, vital organs and bone marrow, so that normal metabolism, DNA synthesis and red blood cell production can take place.

82. In which one of the body fluids is K⁺ concentration higher than that of Na⁺?

(a) Plasma

(b) Perilymph

(d) Cerebrospinal fluid

Ans. (c)

Sol. Endolymph

In scala media

Extracellular fluid

Ionic composition similar to intracellular fluid

High [K⁺], low [Na⁺]

Electric potential of endolymph +80mV = endolymphatic potential

83. Mammillary bodies are present in

(a) thalamus

(b) epithalamus

(d) midbrain

Ans. (c)

Sol. Mammillary Bodies

These are one of the nuclear masses of hypothalamus.

Each contains : medial, lateral and intercalated nuclei.

Recessive fibers from fornix and efferents to antr. thalamus.

Post. to mammiliary bodies, lies postr, performed substance.

84. Reflex ovulation does NOT occur in

(a) cats

(b) rabbits

(d) rats

Ans. (d)

Sol. Reflex ovulation : The group of induced ovulators includes rabbits, ferrets, mink, voles, camels, domestic cats, the short tailed tree shrew and 13-lined ground squirrel. Reflex ovulators are the most efficient mammalian reproducers ovulation and mating are timed accurately. Interestingly coitus hastens the onset and prolongs and surge of LH in some species that ovulate spontaneously. Reflex ovulation occurs in rabbit, cat, ferret, mink and raccoon, while spontaneous or rhythmic ovulation occurs in cow, sheep, mare, bitch, rat and the human.

85. The following statements describe the possible functions of the outer hair cells (OHC) of organ of Corti:

A. The outer hair cells (OHC) are depolarized or hyperpolarized depending on the direction of movement of stereocilia.

B. The OHC are lengthened in depolarization and shortened in hyperpolarization

C. The OHC decrease the amplitude and clarity of sound by shortening and lengthening.

D. The efferent nerve fibers of olivocochlear bundle modulate the sensitivity of the OHC

E. The effect of stimulation of olivocochlear bundle on the OHC is excitatory.

Which one of the options given below represents the correct statements?

(a) A and B

(b) C and D

(d) D and E

Ans. (c)

Sol. In organ of corti, on The outer hair cells (OHC) are depolarized or hyperpolarized depending on the direction of movement of stereocilia and The efferent nerve fibers of olivocochlear bundle modulate the sensitivity of the OHC.

86. The pacemaker cells of sinoatrial node (SA node) are inhibited by the stimulation of vagus nerve. The probable mechanisms of this inhibition are stated as follows:

A. The acetylcholine- regulated K⁺ channels are activated

B. The outward K⁺ causes hyperpolarization of pacemaker cells

C. The inward "funny current" of pacemaker potential is increased

D. The increased intracellular cAMP, induced by activation of M₂ muscarinic receptors, slows the opening of Ca⁺⁺ channels

Choose the answer with correct statements.

(a) A and C

(b) B and C

(d) C and D

Ans. (c)

Sol. The mechanisms of inhibition of pacemaker cells of SA node involve The acetylcholine- regulated K⁺ channels are activated and The outward K⁺ causes hyperpolarization of pacemaker cells.

87. In the glomerular capillary (GC), fluid moves into Bowman's capsule through its almost entire length. But in the muscle capillary (MC), fluid moves into interstitial space at its arteriolar end. The difference between these two capillaries is explained in the following proposed statements:

A. Afferent and efferent arterioles are present on the two ends of GC, but in MC, arteriole and venule are present on two ends.

B. The hydrostatic pressure in GC is higher than that in MC.

C. The efferent arteriole in GC has a relatively low resistance, but venules in MC has a high resistance.

D. The difference of hydrostatic pressure between two ends of GC is relatively more but it is negligible in MC.

E. The difference of oncotic pressure between two ends of MC is negligible but it is relatively more in GC.

F. The net filtration pressure falls to zero at the efferent end of GC but it is 9 mm Hg inward at the venular end of MC.

Which of the above statements are INCORRECT?

(a) A and B

(b) C and D

(d) B and F

Ans. (b)

Sol. Glomerular capillary and muscle capillary are different as Afferent and efferent arterioles are present on the two ends of GC, but in MC, arteriole and venule are present on two ends. The hydrostatic pressure in GC is higher than that in MC. The difference of oncotic pressure between two ends of MC is negligible but it is relatively more in GC and The net filtration pressure falls to zero at the efferent end of GC but it is 9 mm Hg inward at the venular end of MC.

88. Relative rates of red blood cells production in bone marrow of different bones of different ages are shown below :

Identify the correct figure.

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Figure in option (a) correctly represent rate of RBC production in different ages.

89.

The above figure depicts the regulation of hypothalamo-pituitary-adrenal (HPA) axis. Changes in cortisol level in Addison's disease can lead to

(a) suppressed immune system and increased blood glucose level

(b) gain of body weight and lightening of skin

(d) increased blood glucose and activated immune system

Ans. (c)

Sol. Changes in cortisol level in Addison's disease can lead to loss of body weight, reduced blood glucose level and hyperpigmentation.

90. When a skeletal muscle was passively stretched, it contracted reflexly. However, when the muscle was over stretched in this way it showed sudden cessation of contraction followed by relaxation. The following statements provided the possible explanation of these observations:

A. The passive stretching of muscle caused stretching of muscle spindle

B. The over stretching of muscle stimulated Golgi tendon organ

C. Group 1b sensory nerve fibers were stimulated by stretching of muscle spindle

D. Group 1b sensory nerve fibers stimulated -motor neurons that supplied the muscle from which these fibers arose.

E. Group 1a sensory nerve fibers were connected to Golgi tendon organ.

F. The stimulation of group 1a sensory fibers led to the production of IPSP on the ..- motor neurons that supplied the muscle from which these fibers arose.

Which of the above statement(s) is/are correct?

(a) A and B

(b) C and D

(d) C and F

Ans. (a)

Sol. Unlike muscle spindles (located in parallel with muscle fibres), Golgi tendon organs are located in tendons near the myotendinous junction and are in series that is, attached end to end, with extrafusal muscle fibers. Golgi tendon organs are activated when the tendon attached to an active muscle is stretched. As tension in the muscle increases, discharge of the GTOs increases.

The sensory neuron of the GTO synapses with an inhibitory interneuron in the spinal cord, which in turn synapses with and inhibits a motor neuron that serves the same muscle. The result is a reduction in tension within the muscle and tendon.

Thus, whereas spindles facilitate activation of the muscle, neural input from GTOs inhibits muscle activation.

91. Two mutant plants, both bearing white flowers, were crossed, All F₁ plants had red coloured flowers. When an F₁ plant was selfed it produced progeny with either red or white coloured flowers in 9 : 7 ratio. Based on this information, which one of the following conclusions is correct?

(a) The mutations in the parents do not complement each other

(b) The mutations in the parents are allelic

(d) The mutations in the parents are linked

Ans. (c)

Sol. This is an example non allelic complementation : The double heterozygote exhibits a wild-type phenotype, which means that the two mutations are alleles of different genes; each mutant has a functional version of the other gene.

92. In a linkage map, two genes A and B, are 70 cM apart. If individuals heterozygous for both the genes are test crossed number of progeny with parental phenotype will be

(a) equal to the number of progeny with recomninant phenotype

(b) more than the number of progeny with recombinant phenotype

(d) could be more or less than the number of progeny with recombination phenotype depending on whether the genes are linked in cis or trans, respectively

Ans. (a)

Sol. In a linkage map, two genes A and B are 70 cM apart. If individuals heterozygous for both the genes are test crossed number of prog

93. A researcher exposed Drosophila larvae to 37°C during their growth. One of the adult flies that emerged had a crossveinless phenotype. Crossveinless is a known mutant in Drosohila. When this crossveinless fly was crossed to a known crossveinless mutant fly all the progeny had normal phenotype. The observe phenotype can be best explained as an example of

(a) Conditonal mutant

(b) Phenocopy

(d) Pleiotropy

Ans. (b)

Sol. A phenocopy is a variation in phenotype which is caused by environmental conditions, such that the organism's phenotype matches a phenotype which is determined by genetic factors. It is not a type of mutation, as it is non-hereditary.

The larvae of Drosophila melangoaster have been found to be particularly vulnerable to environmental factors which produce phenocopies of known mutations; these factors include temperature, shock, radiation and various chemical compounds. In fruit fly, Drosophila melanogaster, the normal body colour is brownish gray with black margins. Phenocopy can also be observed in Himalayan rabbits. When raised in moderate temperatures. A researcher exposed Drosophila larvae to 37°C during their growth. One of the adult flies that emerged had a crossveinless phenotype. Crossveinless is a known mutant in Drosophila. When this crossveinless fly was crossed to a known crossveinless mutant fly all the progeny had normal phenotype. The observed phenotype can be best explained as an example of Phenocopy.

94. Phages are collected from an infected E. coli donor strain of genotype cys⁺ leu⁺ thr⁺ and used to transduce a recipient of genotype cys^– leu^– thr^–. The treated recipient population is plated on a minimal medium supplemented with leucine and threonine. Many colonies grew. Which one of the following combination of genotypes are appropriate for the colonies that grew?

(a) cys⁺ leu⁺ thr⁺, cys⁺ leu^– thr⁺, cys^– leu⁺ thr^–

(b) cys^– leu⁺ thr⁺, cys⁺ leu^– thr^–, cys⁺ leu⁺ thr^–

(d) cys⁺ leu^– thr^–, cys⁺ leu^– thr⁺, cys⁺ leu⁺ thr^–

Ans. (d)

Sol. Since, medium is supplemented with only leucine and threonine, the recombinants that may grow in such minimal medium would be cys⁺ leu^– thr^–, cys⁺ leu^– thr⁺, cys⁺ leu⁺ thr^–.

95. The inheritance of a given disorder is recorded in three small families shown below:

Based on the above limited information, which one of the following inheritance pattern best explains the observations?

(a) X-linked recessive

(b) X-linked dominant

(d) Autosomal dominant

Ans. (c)

Sol. By observing the pedigree chart the trait would be autosomal recessive.

96. The following table shows mapping data from three interrupted mating experiments using three different Hfr strains and an F. strain:

Appearance of genes in F^– cells

Hfr 1 Genes e⁺ f⁺ c⁺ d⁺ b⁺

Time 6 24 34 49 54

Hfr 2 Genes b⁺ d⁺ c⁺ f⁺ g⁺

Time 1 6 21 31 63

Hfr 3 Genes d⁺ c⁺ f⁺ e⁺ g⁺

Time 4 19 29 47 61

(Time is represented in minutes)

The following answers are derived:

The order of genes is

A. e g b d c f

B. f g b d c e

The distance between

C. f and g is 32 min and between f and b is 30 min

D. c and e is 28 min and between b and c is 20 min

The correct combination of answers is

(a) A, C and D

(b) B and C

(d) B and D

Ans. (a)

Sol. The order of the genes is decided depending on the time taken by the genes to enter the recipient cell. The gene that will be close to site of integration of F plasmid will enter first. Based on that in different strains the genes have entered into recipient cell. By compairing 3 given strains the order of the genes will be egbdcf, the linear order in which thy are entering into recipient cell depending upon the different site of integration of F plasmid in different strains. The time gap in entering two different genes provide distance in minutes which is true for both statements C and D.

97. A new mutant called early ripening (EL) is identified in a plant. The wild type phenotype is late ripening (LR). Further DNA marker(s) is/are observed to be polymorphic between EL and LR plants. A cross was made between pure lines of LR and ER. The F₁ progeny was test crossed and its progeny was analyzed. The parental, F₁ and progeny of test cross were also analyzed for the DNA marker. The table below summarizes the phenotype of the progeny and the pattern of DNA marker observed in each case:

Based on the above information, the following statements were made:

A. LR is dominant to ER

B. The DNA marker used is a dominant marker

C. The DNA marker is linked to the phenotype

Which of the above statements are correct?

(a) A only

(b) A and B only

(d) A, B and C only

Ans. (a)

Sol. By observing presence of DNA marker in parent, F1 progeny and test crossed progeny we can say that LR is dominant to ER.

98. In Burkitt's Lymphoma a reciprocal translocation between chromosome 8 and 14 is observed. If an individual is heterozygous for this translocation, the consequence in meiosis will be as follows:

A. Four chromosomes, i.e., normal chromosome 8 and 14, and translocated chromosome 8 and 14, pair together

B. The two normal chromosomes 8 and 14, and two translocated chromosomes pair separately

C. All gametes produced from this meiosis are non-viable as they have deletions and duplications

D. In one of the cross configurations called "alternate segregation" all gametes having normal or translocated chromosomes, survive

E. The gametes having normal chromosomes only survive while all gametes having translocated chromosomes are non-viable, hence the translocations are used as crossover suppressors

Which of the following combinations best describes the meiotic consequences for the translocation described above?

(a) B and E

(b) B, C and E

(d) A and D

Ans. (d)

Sol. In Burkitt's Lymphoma, four chromosomes, i.e., normal chromosome 8 and 14, and translocated chromosome 8 and 14, pair together and in one of the cross configurations called "alternate segregation" all gametes having normal or translocated chromosomes, survive

99. In Drosophila, Bar eye (B) is a dominant mutation while miniature wing (m) and yellow body colour (y) are recessive mutations. Heterozygous females for these mutations were crossed to normal eyed miniature winged and yellow body coloured males. Assume the following progeny was obtained:

Based on the results obtained, the order of genes will be:

A. B m y

B. m B y

The genetic distance between B and y will be:

C. 40 cM

D. 17.1cM

The correct combination of answer is

(a) A and C

(b) B and C

(d) B and D

Ans. (b)

Sol. Parental combination : B m⁺ y⁺ = 165 and B⁺ m y = 185

Double crossover : B⁺_m⁺ y⁺ = 30 an b m y = 20

Based on parental combination and double crossover order of gene should be m b y.

Single crossover between m and B : B⁺ m⁺ y = 25 and B m y⁺ = 45

Single crossover between B and y : B m⁺ y = 110 and B⁺ m y⁺ = 120

Distance between B and y = = 40 cM

100. Two inbred lines of beans were intercrossed. In F1 the variance in bean length was measured as 2.0. The F1 was selfed to obtain F2 and the variance in bean length in F2 was 7.0. The broad heritability of bean length in the F2 population will be:

(a) 0.75

(b) 9.0

(d) 0.71

Ans. (d)

Sol. The key here is to recognize that all the variance in the F1 population must be environment because all individuals must be of identical genotype. Furthermore, the F2 variance must be a combination of environmental and genetic components, because all the genes that are heterozygous in the F1 will segregrate in the F2 to give an array of different genotypes that relate to bean length. Hence, we can estimate

V_E = 2

V_E + V_G = 7

Therefore, V_G = 7 –2 = 5

Hence, broad heritability (H²) = 5/7 = 0.71

101. Among the following statements, which is the correct one that refers to a Holotype?

(a) Specimens collected from different locations and designated as type specimens by the author

(b) A single specimen or illustration upon which name is based and designated as nomenclature type by the author

(d) A single specimen designated to serve as nomenclatural type when all of the materials on which the name of the taxon was based is missing

Ans. (b)

Sol. A holotype is a single physical example (or illustration) of an organism, known to have been used when the species (or lower-ranked taxon) was formally described. It is either the single such physical example or one of several such, but explicitly designated as the holotype.

For example, the holotype for the butterfly Lycaeides idas longinus is a preserved specimen of that species, held by the Museum of Comparative Zoology at Harvard University. An isotype is a duplicate of the holotype and is often made for plants, where holotype and isotypes are often pieces from the same individual plant or samples from the same gathering.

102. Which one of the following trees represents a fully resolved phylogram?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. It means incomplete separation or a failure to split into two groups unambiguously."There are multiple ways to estimate the resolution of a phylogenetic tree; involving an index to measure the likelihood of observing the phylogenetic tree that you derive relative to those produced by drawing species at random, for instance, or permuting species to produce a null distribution."A badly resolved tree should show up in sets of randomly generated trees far more often than well resolved ones.

103. Identify the plant species from which artemisinin, an anti-malarial drug, is extracted.

(a) Artemisia maritima

(b) Artemisia scoparia

(d) Cinchona officinalis

Ans. (c)

Sol. Artemisisa annua is a medicinal plant whose use has long been reported in China, where it is locally knowns as qinghao. It is now greown commercially in many African countires. A. annua yields artemisnin and the derivatives of this compoound are potent antimalarial drugs. Artemisnin is an endoperoxide sesquiterpene lactone that is effective against multidrug resistant malaria and is also known to act on P. falciparum, the Plasmodium species that causes cerebral malaria.

104. Given below are growth equations where dN/dt is defined as

A. rN/K

B. rN

C. rN[(K-N)/N]

D. rN[(K-N)/K]

With reference to the above equations, which one of the following statements is correct?

(a) B represents exponential growth and A represents logistic growth

(b) B represents exponential growth and D represents logistic growth

(d) A represents exponential growth and D represents logistic growth

Ans. (b)

Sol. Growth curves of population are of the following types :

(a) Exponential growth occurs normally when resources, i.e., food and space are unlimited.

Equation for exponential growth can be

= (b – b) × N

Let (b – d) = r, then

= rN

or N_t = N₀e^rt

Where,

N = Population size

N_t = Population density after time t

N₀ = Population density at time zero

r = Intrinsic rate of natural increase

e = Base of natural logarithms (2.7128)

b = Birth rate and

d = Death rate

Population growth curve. A when responses are not limiting the growth, plot is exponential, B when responses are limiting the growth, plot is logistic, K is carrying capacity.

(b) Logistic growth shows sigmoid curve and this is also called Verhulst-Pearl logistic growth. It can be shown by the following equation.

Where, N = Population density at time t

r = Intrinsic rate of natural increase

K = Carrying capacity

105. Which of the following statements is INCORRECT about a keystone species?

(a) Species other than consumers can be a keystone species

(b) Keystone species has influence on a community proportionate to its abundance

(d) Removing a keystone species can effect successive trophic levels causing a trophic cascade

Ans. (b)

Sol. Keystone species are those species whose importance to an ecosystem's structure, composition and function is disproportionately large relative to their abundance. These species can be of any life form, but they have in common an effect on their environment that is always greater than what can be expected based on their biomass.

A species whose importance to community and ecosystem structure, composition and function is disproportionately large relative to its abundance is referred to as a keystone species. As the name implies, keystone species play key roles in ecosystems. They are disinguishable from dominant species, which also have large roles in ecosystems but solely by virtue of being abundant.

106. A species whose life history strategies allow for high intrinsic rates of increase (rstrategist) will also exhibit the following except

(a) high tolerance for both environmental instability and low quality resources.

(b) short period of exponential population growth (r)

(d) survivorship that show density-dependent mortality, typically exhibiting Type 1 or 2 survivorship curves.

Ans. (d)

Sol. The way a species evolved in and interacts with its niche, it's life history strategy, gives some perspective on a species' means of survival. Most vertebrate animals can be characterized or having a life-history pattern that lies along a continuum from those with a high intrinsic rate of population increase with regular or stochastic fluctuations in population size (r-selected) to those with populations that fluctuate little around a relatively stable environmental carrying capacity.

A species whose life history strategies allow for high intrinsic rates of increase (r-strategist) will also exhibit the reproductive strategy that involves random mating, semelparity and little or no parental investment. It has short period of exponential population growth (r) and high tolerance for both environmental instability and low quality resources.

107. Given below is an ecological pyramid.

The above pyramid represents:

(a) Pyramid of number of a parasitic food chain and pyramid of biomass of a pond ecosystem

(b) Pyramid of number of a pond ecosystem and pyramid of biomass of a forest ecosystem

(d) Pyramid of biomass of a grassland and pyramid of number of a tropical forest ecosystem

Ans. (a)

Sol. Pyramids of numbers : They show the relationship between producers, herbivores and carnivores at successive trophic levels in terms of their number. In a grass land ecosystem the pyramid becomes upright.

Pyramids of numbers : A = grassland ecosystem; B = pond ecosystem; C = Forest ecosystem; D = parasitic food chain; P = producers; PC = primary consumers (herbivores); SC = Secondary consumers (carnivores); TC = Tertiary consumers (carnivores).

Similarly in a pond ecosystem the pyramid is upright. In a forest ecosystem, however, the pyramid of numbers is somewhat different in shape.

The producers, which are mainly large sized trees, are lesser in number and form the base of the pyramid. The herbivores, which are the fruit-eating birds, elephants etc., are more in number than the producers. Then there is a gradual decrease in the number of successive carnivores, thus making the pyramid again upright. However, in a parasitic food chain, the pyramids are always inverted.

108. The rank/abundance plot given below illustrates three well known species abundance curves (a, b, c).

Based on the shapes of the curves select the correct option.

(a) a-exponential, b-log normal, c-geometric series

(b) a-broken stick, b-geometric series, c-log normal

(d) a-exponential, b-geometric series, c-log normal

Ans. (c)

Sol. A rank abundance curve or Whittaker plot is a chart used by ecologists to display relative species abundance, a component of biodiversity. It can also be used to visualize species richness and species evenness. The broken stick corresponds to a community in which all species colonize simultaneously and partition a single resource axis randomly. Log abundance is a measure of a species abundance (e.g., the number of individuals) relative to the abundance of other species.

109. In the context of diversity patterns of species, which one of the following statements is INCORRECT?

(a) Alpha diversity is diversity within a single community

(b) Beta diversity is a measure of the change in species composition from one community or habitat to another

(d) Gamma diversity is the regional diversity found among range of communities/habitats in a geographical region

Ans. (c)

Sol. Alpha divesity was defined by Whittaker as the species richness of a place. However, the practical development of this concept led to redefine the alpha diversity on the basis of the structure of the community. The most common expression is linked, then, both on the number of species and the proportion in which each species is represented in the community.

Beta diversity was defined by Whittaker as "the extent of species replacement or biotic change along environmental gradients." The beta diversity measures the turnover of species between two sites in terms of gain or loss of species.

In ecology, gamma diversity (-diversity) is the total species diversity in a landscape. According to this reasoning, alpha diversity and beta diversity constitute independent components of gamma diversity : Gamma diversity is the regional diversity found among range of communities/habitats in a geographical region.

110. Euphorbiaceae generally represents milky sap bearing plants but there are also some non-milky sap bearing plants that belong to this family. Identify the correct combination of the following given plants which belong to family Euphorbiaceae.

(a) Terminalia bellirica, Euphorbia hirta, Nerium indicum

(b) Mallotus philippense, Ficus religiosa, Ricinus communis

(d) Acalypha indica, Ricinus communis, Mangifera indica

Ans. (c)

Sol. Most spurges such as Euphorbia paralias are herbs, but some, especially in the tropics are shrubs or trees, such as Hevea brasiliensis. The plants of this family are found throughout the world. However, they are not found in arctic regions. In our country the family is represented by several genera such as, Euphorbia, Ricinus, Phyllanthus, Croton, Redilanthus etc.

In the desert regions of Africa and elsewhere the family is represented by cactus like plants of different species of Euphorbia. Mallotus philippense, Acalypha indica and Emblica officinalis belongs to this family.

111. Given below are statements on key characteristics of two fungal lineages.

(A) Microsporidia do not have true mitochondria

(B) Most chytrids produce flagellated gametes

(D) Chytrids reproduce sexually without a dikaryon stage

Choose the combination with correct statements

(a) A, B and C only

(b) A, B and D only

(d) A and B only

Ans. (b)

Sol. Microsporidia are a group of spore-forming unicellular parasites. Microsporidia lack mitochondria, instead possessing mitosomes. They also lack motile structures, such as flagella. Microsporidia produce highly resistant spores, capable of surviving outside their host for up to several years.

Chytrids have flagellated gametes their — reproductive cells have a flagellum that allows them to swim. No other fungi have flagella, which suggests that the other fungi lost this trait at some point in their evolutionary history. This is also consistent with what we know about the closest relatives to the fungi, which also have flagellae.

Most chytrids are unicellular; a few form multicellular organisms and hyphae, which have no septa between cells (coenocytic). They reproduce both sexually and asexually; the asexual spores are called diploid zoospores. Their gametes are the only fungal cells known to have a flagellum.

112. Match the two columns that represent plant organs (I) and parts within these organs (II).

(a) A-(i); B-(iii); C-(ii); D-(iv)

(b) A-(iii); B-(iv); C-(i); D-(ii)

(d) A-(iii); B-(i); C-(ii); D-(iv)

Ans. (d)

Sol. (i) A carpel is the innermost part of a flower. It is usually surrounded by male reproductive structures called stamens, both of which are surrounded by petals. A flower can have one or more carpels. If there are many carpels, they can be separate or fused together. Together, all the carpels are called a gynoecium or a pistil. A carpel is made up of three structures :

Stigma; .Style; Ovary.

(ii) The perianth (perigonium, perigon or perigone) is the non-reproductive part of the flower and structure that forms an envelope surrounding the sexual organs, consisting of the calyx (sepals) and the corolla (petals).

(iii) During the development of sperm cells (SCs, male gamete) from microspores in higher plants, the microspore generated from diploid microsporocytes via meiosis first undergoes asymmetric mitosis to produce a larger vegetative cell (VC) and a smaller generative cell (GC) embedded in the VC. Thereafter, the VC exits the cell cycle and has potential to generate a polarly growing pollen tube; the GC enters further mitosis to produce two SCs for double fertilization.

(iv) The female gametophyte, the megagametophyte, that participates in double fertilization in angiosperms which is haploid is called the embryo sac. This megaspore undergoes three rounds of mitotic division, resulting in seven cells with eight haploid nuclei.

The lower end of the embryo sac consists of the haploid egg cell positioned in the middle of two other haploid cells, called synergids. the synergids function in the attraction and guidance of the pollen tube to the megagametophyte through the micropyle. At the upper end of the megagametophyte are three antipodal cells.

113. Given below are larval stages and phylum to which they belong. Select the INCORRECT combination.

(a) Parenchyma and Amphiblastula - Phylum Porifera

(b) Bipinnaria and Auricularia- Phylum Echinodermata

(d) Planula and Ephyra- Phylum Cnidaria

Ans. (c)

Sol. Axolotl is known as Larva Ambystoma or Tiger Salamander. It has three pairs of outer gills and a caudal fin that bears the tail. Caudal fins are assisting with the swimming process. They belong to the order of the Urodela class of the Amphibians. Tornaria is the planktonic larva of some species of Hemichordata such as the acorn worms. It is very similar in appearance to the bipinnaria larvae of starfishes, with convoluted bands of cilia running around the body. It is an oval shaped, transparent larva.

114. Three species M, N and O when grown independently in a laboratory showed typical logistic growth curves. However, when grown in pairs, the following growth curves were observed.

What interpretation regarding the interspecific relationship between M, N and O can be deduced from the above observations?

(a) N predates over O and therefore can also predate on M.

(b) N is competed out by M and O.

(d) M and O exhibit prey-predator relationship.

Ans. (c)

Sol. In the predator prey relationship, one species is feeding on the other species.

115. During the course of evolution, the jawbones got modified into three ear ossicles in mammals. Which one of the following is a correct match of all three ear ossicles and their jawbones?

(a) stapes – articular; incus – quadrate; malleus – hyomandibular.

(b) stapes – quadrate; incus – articular; malleus – hyomandibular

(d) stapes – hyomandibular; incus – quadrate; malleus – articular

Ans. (d)

Sol. One of the three bones of mammals — stapes, incus and malleus — is really new. The stapes, which began as part of the hyomandibular in the primitive fish, became the single ear bone in amphibians, reptiles and birds. The incus was a prominent upper jawbone (pterygoquadrate) in primitive fish and became reduced to the quadrate in amphibians, reptiles and birds.

It formed a jaw point with the articular in the lower jaw, the latter a modification of Meckel's cartilage.

In the mammal like reptiles the dentary of the lower jaw begins to enlarge; and, finally in the mammals it forms a new jaw joint with the squamosal of the upper part of the skull. Of the two bones of the old joint, the quadrate becomes the incus and the articular becomes the malleus.

116. Temporal isolation in breeding seasons between closely related species leads to reproductive isolation. Given below are breeding seasons of different species of frogs.

Which of the above plots represents temporal isolation in breeding seasons among closely related sympatric species?

(a) Plot A

(b) Plots A and B

(d) Plots A and C

Ans. (c)

Sol. Temporal isolation. Individuals of different species do not mate because they are active at different times of day or in different seasons. Here, in plot B the species P incresases in March to April, whereas Species O is increases in August to November. Similarly same for the plot C.

117. Given below are the population pyramids of three different populations A, B and C depicting the relationship between birth and death rates in each.

Based on the population pyramids given above, which one of the following is INCORRECT?

(a) Population B has slower growth rate than population A.

(b) Population C has birth rate higher than its death rate.

(d) Population B has the highest death rate among the three populations.

Ans. (b)

Sol. By observing the pyramid for population C, the number of individual below the age of 15 years is less than the number of individuals above the age of 65 years hence it contradict the statement 2.

118. Given below are names of some of the National Parks of India and their key protected animals.

Based on the table given above, which of the following options represents the correct match?

(a) A-(iii); B-(i); C-(iv); D-(ii); E-(v)

(b) A-(i); B-(ii); C-(v); D-(iii); E-(iv)

(d) A-(iii); B-(i); C-(v); D-(ii); E-(iv)

Ans. (d)

Sol. National parks in India and IUCN category II protected areas :

Dibru-Saikhowa National Park is a national park in Assam. 36 mammals specise have been recorded of which 12 are listed in Schedule 1 of the Wildlife (Protection) Act of 1972. Species include : Bengal tiger, Indian leopard, clouded leopard, jungle cat, sloth bear, dhole, small Indian civet, Malayan giant squirrel, Chinese pangolin, Gangetic dolphin, slow loris, pig tailed macaque, Assamese macaque, rhesus macaque, capped langur, Hoolock gibbon, Asian elephant, wild boar, Sambar deer, hog deer, barking deer, Asiatic water buffalo and feral horse.

Jaldapara National Park situated in Alipurduar district of West Bengal. Jaldapara forests were constituted as Wildlife Sancturay in the year of 1941 for the protection of Wildlife, particularly one-horned Rhinoceros and it was declared as Jaldapara National Park in May, 2012.

Mukurthi National Park is a 78.46 km² (30.3 sq mi) protected area located in the western corner of the Nilgiris Plateau wes of Ootacamund hill station in the northwest corner of Tamil Nadu. The park was created to protect its keystone species, the Nilgiri tahr.

It is home to any array of endangered wildlife, including royal Bengal tiger and Asian elephant, but its main mammal attraction is the Nilgiri tahr. The park was previously known as Nilgiri Tahr National Park.

Dachigam National Park is located 22 kilometers from Srinagar, Jammu and Kashmir. The name of the park literally stands for "ten villages" which could be in memory of the ten villages that were relocated for its formation.

The Kashmir stag also called hangul is a subspecies of elk native to India. It is found in dense riverine forests in the high valleys and mountains of the Kashmir Valley and northern Chamba district in Himachal Pradesh. In Kashmir, it's found in the Dachigam National Park where it receives protection but elsewhere it is more at risk.

Gir Forest National Park and Wildlife Sanctuary also known as Sasan Gir is a forest and wildlife sanctuary near Talala Gir in Gujarat.

The Asiatic lion is a lion population in Gujarat, which is listed as Endangered on the IUCN Red List because of its small population size. Since 2010, the lion population in and around Gir Forest National Park has steadily increased.

119. Given below are two patterns of ecological succession. Four species are represented by A, B, C and D. An arrow indicates "is replaced by".

In the context of ecological succession, which of the following statements is INCORRECT with respect to the figures given above.

(a) Model – X represents facilitation model and Model - Y represents tolerance model.

(b) Model – X represents tolerance model and Model – Y represents inhibition model.

(d) As per the Model – X, A makes the environments more suitable for B to invade.

Ans. (b)

Sol. X- The facilitation model suggests that the presence of an initial species aids and increases the probability of the growth of a second species. Y- The tolerance model is completely dependent upon life history characteristics. Each species has an equally likely chance to establish itself in the early stages of succession and their establishment results in no environmental changes or impacts on other species. The only possibility for new growth/colonization in this successional sequence arises when a disturbance leads to dominating species being destroyed, damaged, or removed. This frees up resources and allows for the invasion of other species that were not previously present. So, b is incorrect statement.

120. Given below are the species accumulation curves and rarefaction curves measured in an ecological community.

Which one of the following statements is INCORRECT about the two curves?

(a) Species accumulation curve moves from left to right and rarefaction curve moves from right to left.

(b) Species accumulation curve represents the total species richness of the assemblage.

(d) Rarefaction curve is the realized accumulation value of the total species in a community.

Ans. (d)

Sol. A rarefaction is a technique to assess species richness from the results of sampling. Rarefaction allows the calculation of species richness for a given number of individual samples, based on the construction of so-called rarefaction curves.

121. Which one of the following statements is NOT TRUE about the Neutral Theory as proposed by Motoo Kimura?

(a) Except for advantageous mutations, most alleles are under neutral selection

(b) The rate of evolution for most genes will be equal to the neutral mutation rate

(d) At the level of DNA sequences, genetic drift dominates evolution

Ans. (a)

Sol. The neutral theory of molecular evolution (proposed by motto kimura) 1968 holds that at the molecular level most evolutionary changes and most of the variation within and between species is not caused by natural selection but by genetic drift of mutant alleles that are neutral.

Some Important Points :

Except for advantageous mutations most alleles are under neutral condition.

The rate of evolution for most genes will be equal to the neutral mutation rate.

At the level of DNA sequences, genetic drift dominates evolution i.e., genetic drift is the main force changing allele frequency.

The rate of replacement in evolution resulting from the random genetic drift of effectively neutral mutations is equal to the mutations rate to such alleles μ.

Explained the unexpectedly high rate of evolutionary change and very large amount of intraspecific variability at the molecular level that had been uncovered by new techniques in molecular biology.

A neutral mutation is one that does not affect an organism's ability to survive and reproduce.

The neutral theory allows for the possibility that most mutations are deleterious, but holds that because these are rapidly purged by natural selection, they do not make significant contributions to variation within and between species at the molecular level.

Mutations that are not deleterious are assumed to be mostly neutral rather than.

122. Which one of the following statements is NOT TRUE about homologous characters?

(a) Similar traits may not be homologous

(b) Similar traits may be a result of convergent evolution

(d) Homologous characters may show structural similarity but functional diversity

Ans. (c)

Sol. Homologous traits are those traits that are shared by two or more different species that share a common ancestor. These traits are similar in structure or genetics, but may have very different functions and appearances.

Convergent evolution the repeated evolution of similar traits in multiple lineages which all ancestraily lack the trait is rife in nature, as illustrated cause of convergence is usually a similar evolutionary biome, as similar environments will select for similar traits in any species occupying the same ecological niche, enen if those species are only distantly related. In the case of cryptic species.

123. When the electrical response of the two receptors A1 and A2 in a noctuid moth that was exposed to a variety of sounds was measured, it produced the following patterns:

Given this, which one of the following statements is INCORRECT:

(a) The A1 receptor is sensitive to sound of low to high intensity.

(b) A2 receptor begins to produce action potentials only when a sound is loud.

(d) The A1 receptor fires much more frequently to steady, uninterrupted sounds than to high frequency pulses of sound.

Ans. (d)

Sol. "A2 receptor only fires to loud sound. Causes moth to go into powerdive. Observe the graph for high frequency for both the receptor A1 and A2. It clearly shows that there is slight frequency change at high intensity for receptor A1 and A2.

The auditory sensory cells differ by 20-30 dB in noctuid moths and A1 cells are more sensitive than A2 cells.

The A1 cell shows a non-monotonic intensity response function with stimulus durations of 45ms to 1s, while the A2 cell displays a monotomic increase in spike rate with intensity.

The relationship between the increase of A2 cell response and the decrease of A1 cell response is linear and significant".

124. A barn owl sits on its perch 10m above ground. It hears a mouse underneath on the ground at an angle as shown in the figure below.

The error range with which it can locate the mouse on the ground is given by

(a) tan = X/H

(b) cos = X/H

(d) cos = X/C

Ans. (a)

Sol.

125. A 20-week old infant was exposed to the following stimuli and the responses were measured.

Based on the response patterns shown above to the given stimuli, select the correct theory that best describes the observed responses.

(a) Heterogenous summation

(b) Gestalt principle

(d) Sign stimulus

Ans. (b)

Sol. Gestalt psychologists aruged that these principles exist because the mind has an innate disposition to perceive patterns in the stimulus based on certain rules.

These principles are organized into five categories : Proximity, Similarity, Continuity, Closure and Connectedness and given stimuli is based on Gestalt principle.

126. A gene 'X' in Drosophila contributes to inviability of hybrids. The phylogeny below shows the evolutionary history of gene 'X' and on each branch the numbers indicate the non-synonymous/synonymous substitutions.

Based on the above information, select the correct statement from the choices below:

(a) High proportion of non-synonymous changes at A indicates evolution by natural selection.

(b) High proportion of synonymous changes at A indicates absence of natural selection.

(d) Low proportion of non-synonymous changes at C indicates positive selection.

Ans. (a)

Sol. The rate of evolution in protein-coding genes is commonly assessed with the two quantities dN (rate of nonsynonymous substitutions per nonsynonymous site, also called Ka) and dS (rate of synonymous substitutions per synonymous site, also called Ks).

If synonymous evolution in neutral, then the ratio of dN/dS identifies the type of selection pressure acting on a gene. dN/dS 1 indicates strong purifying selection, dN/dS 1 indicates positive selection and dN/dS = 1 implies that amino acids evolve largely neutrally. At C 3/10 = dN/dS = 0.3 which is less than 1 hence it indicates positive selection.

127. When James R Brown and W Ford Doolittle (1997) reconstructed the tree of life using a variety of different genes, they found that different genes gave fundamentally different phylogenies as shown below. Note that the genes are unique to the specific trees.

From the given options select the process that best explains the observed discrepancies between the trees.

(a) polyploidization

(b) horizontal gene transfer

(d) localized extinctions

Ans. (b)

Sol. Horizontal gene transfer (HGT) or lateral gene transfer (LGT) is the movement of genetic material between unicellular and/or multicellular organisms other than by the ("vertical") transmission of DNA from parent to offspring.

The phylogenetic trees to genes are called 'gene trees'. Reconstruction of gene trees is quite important for evolutionary studies, because replication of nucleotide sequences automatically produces a bifurcating tree of genes.

When James R Brown and W Ford Doolittle (1997) reconstructed the tree of life using a variety of different genes, they found that different genes gave fundamentally different phylogenies.

In contrast to vertical gene transfer from parent to offspring, horizontal (or lateral) gene transfer moves genetic information between different species.

Bacteria and archaea often adapt through horizontally gene transfer. Recent analyses indicate the eukaryotic genomes, too, have acquired numerous genes via horizontal transfer from prokaryotes and other lineages.

128. Orchids of the genus Cryptostylis are known to maintain reproductive isolation because their flowers look and smell like females of the wasps of genus Lissopimpla. When the male wasp visits and attempts to mate with the flower, the shape of anther and stigma allows correct placement and transfer of pollen to the wasp, which then transfers the pollen to species specific flower that it next attempts to mate with. This prezygotic barrier that prevents inter-species cross-pollination in Cryptostylis is best explained by:

(a) behavioural isolation through mimicry

(b) mechanical isolation through mimicry

(d) habitat isolation

Ans. (b)

Sol. Mechanical isolation

Difference in size and shape of reproductive organs makes mating impossible.

In plants such as orchids, mechanical isolation may involve pollinators.

Mechanical Isolation through Mimicry Temporal Isolation

Mating periods do not overlap.

In sympatric populations of threee closely related leopard frogs, each species breeds at a different time of year.

In the Ophrys type, floral scents attract male bees or wasps and play a role in their mating behaviour; different species of flowers, often orchids, have different scents and attract different sets of hymenopterana species.

129. Which one of the following microscopes would you use to visualize a protein fused to an appropriate reporter in a living cell?

(a) Fluorescence microscope

(b) Scanning electron microscope

(d) Phase contrast microscope

Ans. (a)

Sol. Fluorescence microscopy with fluorescent reporter proteins has enabled analysis of live cells by fluorescence microscopy, however cells are susceptible to phototoxicity, particularly with short wavelength light. Furthermore, fluorescent molecules have a tendency to generate reactive chemical species when under illumination which enhances the phototoxic effect.

130. Which one of the following statements regarding use of hybrid nucleases for plant genome engineering is INCORRECT?

(a) For gene knock out experiments, the nuclease is expressed without a donor DNA template.

(b) Hybrid nucleases typically comprise two protein subunits that dimerize at their nuclease domain.

(d) Both ZFNs and TALENs are fused with cleavage domains of FokI endonuclease.

Ans. (c)

Sol. Targeted genome editing in plants using engineered nucleases ('GEEN'). Engineered nucleases are used to induce a double-stranded DNA break (DSB) at a specified locus of the gene of interest (GOI). DSBs are repaired by either non-homologous end-joining (NHEJ) or homologous recombination. Zinc-finger nucleases are artificial restriction enzymes generated by fusing a zinc finger DNA-binding domain to a DNA-cleavage domain. Zinc finger domains can be engineered to target specific desired DNA sequences and this enables zinc-finger nucleases to target unique sequences within complex genomes.

131. Which of the following host systems is best suited to express large amounts of glycosylated protein for structural studies?

(a) GST fusion protein in E. coli

(b) His-tagged protein in E. coli

(d) Native protein in Pseudomonas fluorescens

Ans. (c)

Sol. Baculoviridae is a family of viruses. Arthoropods, lepidoptera, hymenoptera, dipetra and decapoda serve as natural hosts. Baculoviruses are known to infect invertebrates, with over 600 host common hosts, but these viruses have also been found infecting sawflies, mosquitoes and shrimp.

Polyhedrin negative baculovirus expression system is susceptible to desiccation and UV light; survival time is limited to hours. Native protein in baculovirus is best suited to express large amounts of glycosylated protein for structural studies.

132. Which one of the following statements regarding proteins is CORRECT?

(a) transition requires less energy than transition and can be monitored by mass spectrometry.

(b) transition requires more energy than transition and can be monitored by UV-Vis spectroscopy.

(d) transition requires less energy than transition and can be monitored by CD spectroscopy.

Ans. (c)

Sol. n to star transition () involves the least amount of energy than all types of transition in ultraviolet visible spectroscopy. Therefore, the transition gives the absorption with a longer wavelength.

133. Which of the following statements about YACs is FALSE?

(a) YACs can carry DNA fragments of 1Mb

(b) A single yeast may contain more than one YAC

(d) YACs have to be transferred to bacteria for subsequent DNA manipulations

Ans. (c)

Sol. Yeat artificial chromosome (YAC) and bacterial artificial chromosome (BAC) are two types of vectors involved in cloning. YAC is an artifically constructed vector system using a specific region of yeast chromosome to insert large segments of genetic materials to yeast cells. Technological improvements now make the cloning of large DNA pieces possible, using artificially constructed chromosome vectors that carry human DNA fragments as large as 1 Mb. These vectors are maintained in yeast cells as artificial chromosomes (YACs).

YAC (Yeast artificial chromosome) is an artifically built chromosome which has the ability to carry a large segment of foreign DNA and replicate within the yeast cells. Yeast recombination is very feasible and always remains active. Hence it can generate deletions and other rearrangements in a YAC. It have to be transferred to bacteria for subsequent DNA manipulations.

134. In an intact cell patch-clamp experiment,

(a) two electrodes are inserted into the cytosol but at different depths.

(b) one electrode is applied to the plasma membrane in a region containing only lipids and one into the cytosol.

(c) two electrodes are applied to the plasma membrane, one in a region containing only lipids and the other in a region containing one or few ion channels.

(d) one electrode is applied to the plasma membrane containing one or few ion channels and one electrode inserted into the cytosol.

Ans.

Sol. This patch is small enough that it will usually contain only one or perhaps a few ion channels. Current can enter and leave the pipette only through these channels, thereby enabling an experimenter to study various properties of the individual channels.

The channels can be studied in the intact cell or the patch can be pulled away from the cell so that the researcher has access to the cytosolic side of the membranes.

135. Identification of genes that are associated with the development of male and/or female gametophyte and embryogenesis in plants is facilitated by T-DNA mediated insertional mutagenesis. In an experiment, a transgenic plant was generated by insertion of T-DNA (containing a Kanamycin-resistance gene) into a gene "A". Self pollination of the T0 plant generated F1 progeny that segregated in a 2:1 ratio for resistance:sensitivity to Kanamycin. These observations indicate that

(a) the mutant allele did not segregate from the wild type allele.

(b) mutation in gene "A" induces lethality in the male gametophyte.

(d) mutation in gene "A" induces zygotic lethality.

Ans. (d)

Sol. The zygote would normally develop into an embryo, as instructed by the genetic material within the unified cell. However, a zygotic lethal gene scotches prenatal development at its earliest point. A zygotic lethal gene is a mutated (changed) version of a normal gene essential to the survival of the zygote. So mutation in gene "A" by insertional mutagenesis induce zygotic lethality .

136. Some of the following transgenic approaches could be used for functional characterization of endogenous genes in plants:

A. Transformation using a binary vector containing a strong enhancer element and lacking the right border of T-DNA

B. Transformation using a binary vector containing a promoter-less reporter gene sequence and a selection maker gene cassette within the T-DNA.

C. Transformation using a binary vector containing only a strong enhancer element and a selection marker gene cassette within the T-DNA.

D. Transformation using a binary vector lacking a reporter gene as well as both the left and right borders of T-DNA.

Which one of the following combinations can be used?

(a) A and B only

(b) B and C only

(d) A and D only

Ans. (b)

Sol. Functional characterization of an endogenous gene can be done by overexpressing that gene of interest with the help of A. tumefaciens. To accomplish this, the gene of interest has to be inserted within the T- DNA region so that it can be successfully transferred to plants. TDNA-regions are defined by T-DNA border sequences. These borders are 25 bp in length and highly homologous in sequence. They flank the T-region in a directly repeated orientation. In general, the T-DNA borders delimit the T-DNA, because these sequences are the target of the VirD1/VirD2 border-specific endonuclease that processes the T-DNA from the Ti plasmid. There appears to be a polarity established among T-DNA borders: right borders initially appeared to be more important than left borders. border sequences not only serve as the target for the VirD1/VirD2 endonuclease but also serve as the covalent attachment site for VirD2 protein. Therefore left and right borders of T-DNA are crucial for transfer of genes.. In addition to that selection marker is essential for screening transgenic plants. So, Option A and D are incorrect. Option A and D says the TDNA that is transferred has to be either without one or both the borders which means gene transfer will not happen in both the cases. therefore only option (b) and (c) are correct.

137. Match the following Vir proteins with their correct function during Agrobacterium mediated transfer of T-DNA to plant cells.

(a) A-(i), B-(ii), C-(iii), D-(iv)

(b) A-(iv), B-(i), C-(ii), D-(iii)

(d) A-(iii), B-(ii), C-(i), D-(iv)

Ans. (b)

Sol. 1. A set of Agrobacterium tumefaciens operons required for pathogenesis is corrdinately induced during plant infection by the VirA and VirG proteins.

2. VirD2 is one of the key Agrobacterium tumefaciens proteins involved in T-DNA processing and transfer. In addition to its endonuclease domain, VirD2 contains a bipartile C-terminal nuclear localization sequence (NLS) and a conserved region called omega that is important for virulence.

3. VirB1, a component of the T-complex transfer machinery of Agrobacterium tumefaciens, is processedd to a C-terminal secreted product, VirB1.

4. Some additional proteins like importins, VIP1 and VirF may interact with the T-strand, either directly or indirectly to form larger T-complexes in the plant cell. VirF directs the proteins coating T complex (VIP1 and VirE2) for destruction in proteasome.

138. Match the technique with its appropriate application/use from the list of options given below

(a) A-(iii), B-(i), C-(iv), D-(ii)

(b) A-(ii), B-(iii), C-(i), D-(iv)

(d) A-(iii), B-(iv), C-(ii), D-(i)

Ans. (a)

Sol. 1. Chromatin immunoprecipitation coupled to next generation sequencing (ChIP-seq) is widely used to study the in vivo binding sites of transcription factors (TFs) and their regulatory targets. Recent improvements to ChIP-seq, such as increased resolution, promise deeper insights into transcriptional regulation, yet require novel computational tools to fully leverge their advantages.

2. Bisulfite sequencing (also known as bisulphite sequencing) is the use of bisulfite treatment of DNA to determine its pattern of methylation. DNA methylation was the first discovered epigenetic mark and remains the most studied. In animals it predominantly involves the addition of a methyl group to the carbon-5 position of cytosine residues of the dinucleotide CpG and is implicated in repression of transcriptional activity.

3. Enzyme Linked Immunosorbant Assay (ELISA) kit, modified the protocol for application to plant tissues and used it to quantify levels of NPTII protein in transformed plants.

The ELISA proved safe, economical and convenient to reliably screen and quantify NPTII protein in large numbers of plant samples.

The sensitivity of the ELISA for NPTII detection in tobacco plants is at least an order of magnitude greater thana widely used radioactive gel assay.

4. RAPID amplification of cDNA ends (RACE) is a technique used in moelcular biology to obtain the full length sequence of an RNA transcript found within a cell. RACE results in the production of a cDNA copy of the RNA sequence of interest, produceed through reverse transcription, followed by PCR amplification of the cDNA copies. RACE can provide the sequence of an RNA transcript from a small known sequence within the transcript to the 5' end (5' race-pcr) or 3' end (3' race-pcr) of the RNA.

This technique is sometimes called one-sided PCR or anchored PCR.

139. To achieve a best resolution using a fluorescence microscope, what combination of wavelength of emitted light (), refractive index and the angle (2) by which light enters into the microscope would be the best choice for the user:

(a) =405; refractive index=1.33; 2=90.

(b) =420; refractive index=1.51; 2=180.

(d) =405; refractive index=1.51; 2=180.

Ans. (d)

Sol. Resolution = 0.61 /n sin θ

Maximum resolution in achieved at = 90,

so 2 = 180

N = 1.51

so n sin = 1.51 sin 90 = 1.51

At = 405 nm

r = 0.61 × 405/1.51

= 163.60

at = 0.61 × 420/1.51

= 169.66

so r is the smallest resolvable distance. Lower the value of r, higher the resolution.

140. A researcher attempted to clone a 630bp coding sequence of a gene downstream to a bacterial promoter (500bp in length) for over expression and purification of the encoded protein. The gene sequence was isolated as a SmaI fragment (Recognition sequence: arrow indicates the site of restriction) and cloned at a SmaI site located downstream to the promoter. The gene sequence contained a single site for EcoRI located 30bp downstream to the start codon. A schematic representation of the plasmid along with locations(s) of some restriction enzyme sites of the vector is given below.

The researcher screened the obtained colonies by a double digestion using HindIII and EcoRI. Which one of the following restriction digestion patterns would represent the restriction profile of the desired clone that could be used for overexpression?

(a) ~530bp + ~600bp + vector backbone

(b) ~1100bp + ~30bp + vector backbone

(d) ~1130bp + vector backbone

Ans. (a)

Sol. Restriction digestion pattern would be of ~530bp due to restriction digestion by Hind3 and EcoR1 (500+30) + ~600bp + vector backbone.

141. Following observations are made regarding a peptide sequence.

— The peptide is inert to Ellman's reagent. However, on reacting with β-mercaptoethanol, the peptide gives a positive Ellman's test.

— The peptide sequence gives a broad minimum around 211 nm in the CD spectrum.

— With increasing concentration of the peptide, the melting temperature of the peptide increases.

— On treating the peptide with D₂O, half the total number of amides get exchanged.

Which one of the following statements is correct?

(a) It is an -helical peptide that undergoes aggregation

(b) It is an -helical disulfide bridged peptide that undergoes aggregation

(d) The peptide is composed of an -helix and -sheet connected by a disulfide bridge

Ans. (c)

Sol. Southern blot and electron microscopy are not common.

142. Given below are spatial and temporal techniques (column I) used to detect brain activity (column II)

Select the correct set of combination

(a) A-(i), B-(ii), C-(iii), D-(iv)

(b) A-(iii), B-(i), C-(iv), D-(ii)

(d) A-(ii), B-(iv), C-(i), D-(iii)

Ans. (b)

Sol. 1. Functional magentic resonance imaging or fMRI, is a technique for measuring brain activity. It works by detecting the changes in blood oxygenation and flow that occur in response to neural activity when a brain area is more active it consumes more oxygen and to meet this increased demand blood flow increases to the active area. fMRI can be used to produce activation maps showing which parts of the brain are invovled in a particular mental process.

2. Positron Emission Tomography (PET) uses trace amount of short-lived radioactive material to map functional processes in the brain.

When the material undergoes radioactive decay a positron is emitted, which can be picked up be the dectector. Areas of high radioactivity are associated with brain activity.

3. Computed tomography (CT) scanning builds up a picture of the brain based on the differential absorption of X-rays. During a CT scan the subject lies on a table that slides in and out of a hollow, cylindrical apparatus. An x-ray source rides on a ring around the inside of the tube, with its beam aimed at the subjects head.

After passing through the head, the beam is sampled by one of the many detectors that line the machine's circumference.

4. Electroencephalography (ECG) is the measurement of the electrical activity of the brain by recording from electrodes placed on the scalp.

The resulting traces are known as an electroencephalogram (EEG) and represent an electrical signal from a large number of neurons.

143. If a metabolically active cell is challenged with ⁵⁵Fe radioisotope label, which of the following proteins CANNOT be detected by autoradiography?

(a) Aconitase and lipoic acid synthase

(b) Cytochrome C and DNA primase

(d) Myoglobin and Homoaconitase

Ans. (c)

Sol. Calcineurin is a calmodulin-binding protein in brain and the only serine/threonine protein phosphatase under the control of Ca²⁺/calmodulin (CaM), which plays a critical role in coupling Ca²⁺ signals to cellular responses. CaM up-regulates the phosphatase activity of calcineurin by binding to the CaM-binding domain (CBD) of calcineurin subunit . Any of the Calmodulin and Calcineurin don't have Fe atom , so they cannot be detected by autoradiography using Fe radioisotope model.

144. Diagnosis of influenza virus infections can be done using some of the following techniques:

A. Western blot and Southern blot

B. Northern blot and western blot

C. ELISA and RT-PCR

D. PCR and electron microscopy

Choose the combination of techniques that correctly lists the detection methods.

(a) A and B only

(b) C and D only

(d) A and D only

Ans. (c)

Sol. Despite significant advancement in vaccine and virus research, influenza continues to be a major public health concern.

Each year in the United States of America, influenza viruses are responsible for seasonal epidermis resulting in over 200,000 hospitalizations and 30,000-50,000 deaths. Serological assays most commonly used to detect influencza virus-specific antibody responses include hemaggluti-nation inhibition assay (HAI), microneutralization or virus neutralization assay (VN), single radial hemolysis (SRH), complement fixation assay, enzyme linked immuno-absorbant assay (ELISA) and Western blotting.

145. From the following statements:

(A) Coloured images can be obtained by transmission electron microscopy by fluorescent labelling of the specimen

(B) Scanning electron microscopy requires sectioning of the sample

(C) Confocal microscopy uses optical methods to obtain images from a specific focal plane and excludes light from other planes

(D) Differential-interference microscopy relies on interference between polarized light due to differences in the refractive index of the object and surrounding medium

(E) Visualization in epifluorescence microscopy requires staining by heavy metal atoms

Choose the combination with two correct and one incorrect statements.

(a) B, C, E

(b) A, B, E

(d) B, D, E

Ans. (c)

Sol. Two different freeze-fracture methods are explored for preparation of biological material for scanning electron microscopy. In the simpler method the tissues are first fixed and dehydrated. They are then frozen and fractured. In the second method tissues or cells are first infiltrated with cryprotectant (dimethylsulphoxide) then frozen and fractured and not fixed until after thawing. The fixed tissues are finally dehydrated and critical point dried and after thawing, criticap point dried. Confocal microscopy, most frequently confocal laser scanning microscopy (CLSM) is an optical imaging technique for increasing optical resolution and contrast of a micrograph by means of using a spatial pinhole to block out-of-focus light in image formation. Therefore B and C are right.

In order for a sample to be suitable for fluorescence microscopy it must be fluoresenent. There are several methods of creating a fluorescent sample; the main techniques are labelling with fluorescent stains or in the case of biological samples, expression of a fluorescent protein. So E is wrong.

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Previous Year Question Paper with Solution.