PYQ JUNE 2016 - VedPrep

CSIR NET BIOLOGY (JUNE - 2016)
Previous Year Question Paper with Solution.

21. The solubility of gases in water depends on their interaction with water molecules. Four gases i.e. carbon dioxide, oxygen, sulphur dioxide and ammonia are dissolved in water. In terms of their solubility which of the following statement is correct?

(a) Ammonia > Oxygen > Sulphur dioxide > Carbon dioxide

(b) Oxygen > Carbon dioxide > Sulphur dioxide > Ammonia

(d) Ammonia > Sulphur dioxide > Carbon dioxide > Oxygen

Ans. (d)

Sol. Most soluble gas in water is ammonia (NH₃). The reason of poor solubility of O₂ or N₂ or H₂ in water is that they just occupy the intermolecular spaces and stays there by interaction with its surrounding molecules by weak van der Waals forces (H₂, N₂ or Ar etc) or very weak electronic interactions with by use of its pi electrons (C₂H₂) or Ione pairs (like O₂) with water, so O₂ has higher solubility than N₂ in water.

CO₂ when dissolves in water it interacts with surrounding H₂O molecules to hydrate itself, i.e., the electronegative oxygen atoms of CO₂ forms strong H-bond with H atoms of water molecules. This interaction is very strong to dissolve more and more CO₂ molecules.

Hence, sequence according to increasing order of solubility of gases in water is like this :

Oxygen < Carbon dioxide < Sulphur dioxide < Ammonia.

22. Penicillin acts as a suicide substrate. Which one of the following steps of catalysis does a suicide inhibitor affect?

(a) k₁

(b) k₂

(d) k₄

Ans. (c)

Sol. Suicide inhibition is a form of irreversible inhibition in which the substrate in the first catalytic cycle is converted into a chemically reactive product which remains bound to the active site through covalent bonding. The enzyme is rendered permanently inactive. Such inhibitors have potential as drugs.

23. Which of the following is not true for cholesterol metabolism?

(a) HMG-CoA reductase is the key regulator of cholesterol biosynthesis.

(b) Biosynthesis takes place in the cytoplasm.

(d) Cholesterol is transplanted by LDL in plasma.

Ans. (c)

Sol. Cholesterol is an extremely important biological moelcule that has roles in membrane structure as well as being a precursor for the synthesis of the steroid hormones, the bile acids and vitamin D.

HMG-CoA reductase and all subsequent enzymes involved in cholesterol biosynthesis are associated with the endoplasmic biosynthesis are associated with the endoplasmic reticulum and have also been reported to occur in rat liver peroxisomal membranes.

In Cholesterol biosynthesis of nucleotides, protein and fatty acids takes place in cytoplasm.

It may well be that HDL makes a larger contribution to cell cholesterol in general in those species (such as the rat) in which HDL is the major carrier of plasma cholesterol than in species (such as man and the pig) in which most of the plasma cholesterol is carried in LDL

24. Predominant interactions between phospholipids that stabilize a biological membrane include

(a) hydrogen bonds and covalent interactions.

(b) van der Waals and ionic interactions.

(d) covalent and hydrophobic interactions.

Ans. (c)

Sol. The Cytoplasmic Membranes (CM) is stabilized by hydrophobic interactions and hydrogen bonds; the former are also known as van der Waals interactions. In addition, divalent cations such as Mg²⁺ and Ca²⁺ stabilize the membrane by neutralizing the negative charges of the phospholipids on both sides of the bilayer, serving as 'salt bridges'.

25. The –COOH group of cellular amino acids can form which of the following bonds inside the cell?

(a) Ether and ester bonds

(b) Ester and amide bonds

(d) Amide and carboxylic anhydride bonds

Ans. (b)

Sol. Two of the most common linkages between functional groups are ester bonds, which form between carboxylic acids and alcohols and amide bonds, which form between carboxylic acids and amines.

26. The standard free energy change per mole for the reaction A B at 30°C in an open system is –1000 cal/mole. What is the approximate free energy when the concentration of A and B are 100 micromolar and 100 millimolar, rerspectively?

(a) 3160

(b) 316

(d) –3160

Ans. (a)

Sol. + 2.303 RT = –1000 cal/mol + 2.303 × 2 × 303

= –1000 cal/mol + 2.303 × 2 × 303 × log 10³ = –1000 cal/mol + 4186.854 cal/mol = 3186.8654 cal/mol

27. Indicate which one of the following statements about nucleic acids and protein structures is correct?

(a) Hydrogen bonding between the bases in the major and minor grooves of DNA is absent.

(b) Both uraci and thymine have a methyl group but at different positions.

(c) The backbone dihedral angles of -helices and -sheets are very similar. Only the hydrogen bonding pattern is different.

(d) A -turn is formed by four amino acids. The type of -turn is determined by the dihedral angles of the second and third amino acid.

Ans. (d)

Sol. Beta turns (-turns, -bend, tight turns, reverse turns) are very common motifs in proteins and polypeptides. Each consists of four amino acid residues.

The four types of beta turn are distinguished by the angles of residues i + 1 and i + 2 as shown in the table below giving the typical average values. Glycines are especially common at amino aicds with positive angles; for prolines such a conformation is sterically impossible but they occur frequently at amino acid positions where is negative.

The main chain atoms of type I and I' turns are enantiomers (mirror images) of one another. So are the main chain atoms of type II and II' turns.

28. The turnover number and specific activity of an enzyme (molecular weight 40,000 D) in a reaction (V_max = 4 µmol of substrate reacted/min, enzyme amount = 2 µg) are

(a) 80,000/min, 2 × 10³ µmol substrate/min

(b) 80,000/min, 2 × 10³µmol substrate/second

(d) 40,000/min, 2 × 10³ µmol substrate/min

Ans. (a)

Sol. E_t = µmol enzyme

V_max = 4 µmol of substrate reacted/min

Specific activity = = 2 × 10³ µmol substrate/min

However, the standard unit of specific activity is µmol substrate min^–1 mg^–1 i.e. total activity per mg protein present in the solution.

29. A researcher has developed a program to evaluate the stability of a protein by substituting each amino acid at a time by the other 19 amino acids. For a protein, a researcher has observed the following changes in stability upon substitution of amino acids in loops, helices, sheets, protein core and on the protein surface.

Substitutions in

1. loops are more tolerant

2. sheets are more tolerant

3. core is less tolerant

4. helices are less tolerant

5. surface is more tolerant

Which of the above statements are correct?

(a) 1 and 3

(b) 3 and 4

(d) 1 and 2

Ans. (a)

Sol. Substitution of amino acids in loops is more tolerant to stability of protein due to change in shape of loops. While core is less tolerant to substitution of amino acid.

30. Lateral diffusion of proteins in membrane can be followed and diffusion rate calculated by

(a) Atomic force microscopy

(b) Scanning electon microscopy

(d) FRAP

Ans. (d)

Sol. The movement of protein (and lipids) in the plane of bilayer can be demonstrated by fluorescence microphotolysis or fluoresence recovery after photobleaching (FRAP). In FRAP, cell is made to fluoresce by attaching a fluorescent probe to the molecules moving in the plane of the membrane. A small patch of membrane is then irradiated (bleached) and the rate of increase in fluorescence in the bleached area over a period of time (recovery of fluorescence) is measured to calculate the diffusion coefficient (D) of the molecule. Diffusion coefficient of several membrane proteins and lipids has been measured. The diffusion coefficient of rhodopsin (5 × 10^–9 cm²/s) has been found to be highest of any known protein. The rate of diffusion of glycoproteins in membrane is usually 5-50 times less than that of rhodopsin.

31. Labelling of membrane spanning domain of any integral membrane protein in a given plasma membrane vesicle (without disrupting its structure) is successfully carried out by

(a) immunochemical methods

(b) metabolic labelling with radioisotopes

(d) limited proteolysis followed by metabolic labelling

Ans. (c)

Sol. Photoaffinity labeling is a chemoproteomics technique used to attach "labels" to the active site of a large molecule, especially a protein. Photoaffinity labeling (PAL) using a chemical probe to covalently bind its target in response to activation by light has become a frequently used tool in drug discovery for identifying new drug targets and molecular interactions, and for probing the location and structure of binding sites.

32. Both sphingomyelin and phosphoglycerides aer phospholipids. Which one of the following statements is not correct?

(a) While one has a fatty acid tail attached via an ester bond, in another, the fatty acid tail is attached via an amide bond.

(b) The hydrophilicity of both is dependent on the phosphate group and other head groups attached to the phosphate group.

(d) Both may have choline as head group.

Ans. (c)

Sol. Phospholipids : Phospholipids and glycolipids consist of two long, nonpolar (hydrophobic) hydrocarbon chains linked to a hydrophilic head group.

The heads of phospholipids are phosphorylated and they consist of either :

Glycerol (and hence the name phosphoglycerides given to this group of lipids) or

Sphingosine (e.g., sphingomyelin and ceramide).

Phosphoglycerides : In phosphoglycerides, the hydroxyl groups at C-1 and C-2 of glycerol are esterified to the carboxyl groups of the FAs. The C-3 hydroxyl group is esterified to phosphoric acid. The resulting compound, called phosphatidate, is thee simplest phosphoglycerate. Only small amounts of phosphatidate are present in membranes. However, it is a key intermediate in the biosynthesis of the other phosphoglycerides.

Sphingolipids : Sphingosine is an amino alcohol that contains a long, unsaturated hydrocarbon chain. In sphingomyelin and glycolipids, the amino group of sphingosine is linked to FAs by an amide bond. In sphingomyelin the primary hydroxyl group of sphingosine is esterified to phosphoryl choline.

In glycolipids, the sugar component is attached to this group. The simplest glycolipid is cerebroside, in which there is only one sugar residue, either Glc or Gal. More complex glycolipids, such as gangliosides contain a branched chain of as many as seven sugar residues.

33. It is well established that "Band 3" protein of red blood cell membrane is solely responsible for Cl^– binding site of "Band 3" is crucial for this event. Keeping this in mind what is the most appropriate way to load and retain a small anionic fluorescent probe (x) inside the red blood cells (RBCs) suspended in phosphate bufferred saline (PBS), pH 7.4?

(a) Incubate the RBCs with x in phosphate buffered saline (PBS, pH 7.4) at 37°C for 30 min.

(b) Incubate the RBCs with x in PBS at 4°C for 30 min.

(d) Incubate the RBCs with x in Hepes sulphate buffer (pH 7.4) at 37°C for 30 min. followed by treatment with a NH₂ group modifying agent (covalent modification).

Ans. (d)

Sol. Specifically, HEPES buffer is used to maintain the ideal conditions for cell culture. Therefore, it's classed as a biological buffer and is best suited for microbiology and histology. It's also useful in medicine as part of diagnostic laboratory procedures.

34. Histone deacetylase (HDAC) catalyses the removal of acetyl group from N-termina of histones. Which amino acid of histone is involved in this process?

(a) Lysine

(b) Aginine

(d) Histidine

Ans. (a)

Sol. Histone deacetylase (HDAC) catalyses the removal of acetyl group from N-termina of histones. Acetylation as well as Deacetylation occurs for lysine amino acid.

35. E. coli is being grown in a medium containing both glucose and galactose. On depletion of galactose, experssion of -galactoside wil

(a) remain unchanged

(b) increase

(d) initially decrease and then incerase

Ans.

Sol. When E. coli finds both glucose and lactose in the medium, it preferentially uses the glucose and the use of lactose is prevented until the glucose is used up, causing a biphasic growth. On depletion of glucose, expression of -galactoside will increases.

36. Error-free repair of double strand breaks in DNA is accomplished by

(a) non-homologous end-joining

(b) base excision repair

(d) mismatch repair

Ans. (c)

Sol. In lower eukaryotes such as yeast, homologous recombination repair is the predominant pathway used for repairing DNA double-strand breaks. Double-strand break repair by homologous recombination is an error-free process because repair is performed by copying information from the undamaged homologous chromatid/chromosome.

37. RNA interference is mediated by both siRNA and miRNA. Which one of the following statement about them is not true?

(a) Both siRNA and miRNA are processed by DICER.

(b) Both siRNA and miRNA usually guide silencing of the same genetic loci from which they originate.

(d) miRNA, but not siRNA is processed by Drosha.

Ans. (b)

Sol. An important distinction between miRNAs and siRNAs is whether or not they silence their own expression. Almost all siRNAs silence the same locus as they were derived from and they only sometimes have the ability to silence other loci as well. In contrast, most miRNAs do not silence their own loci but do silence other genes. Both RNAs have double-stranded precursors and depend upon the same two families of proteins : Dicer enzymes to excise them from their precursors and Ago proteins to support their silencing effector functions.

38. E. coli was grown in three different experimental conditions. In one, it was grown in medium containing glucose as carbon source; in the second in medium containing both glucose and galactose; and in third was infected with phage. Match the curves shown below to the treatment.

(a) P is grown in glucose; Q is grown in glucose and galactose; R is infected with phage

(b) P is grown in glucose and galactose; Q is grown in glucose; R is infected with phage

(d) P is infected with phage; Q is grown in glucose; R in glucose and galactose

Ans. (b)

Sol. By observing the curves, it can be assumed that P is grown in glucose and galactose; Q is grown in glucose; R is infected with phage.

39. Microsatellites are used as marker for identifying individuals via DNA fingerprinting as the alleles may differ in the number of repeats. From the Southern blot shown below identify the progeny (A, B, C and D) for the given parents (M = mother, F = father).

(a) A, B, C and D

(b) A, B and D

(d) B, C and D

Ans. (b)

Sol. By observing positions of micro-satellite markers A, B and D are progeny of M and F.

40. Each amino acyl-tRNA synthetase is precisely able to match an amino acid with the tRNA containing the correct corresponding anticodon. Most organisms have 20 different tRNA synthetases, however some bacteria lack the synthetase for charging the tRNA for glutamine (tRNA^Gln) with its cognate amino acid. How do these bacteria manage to incorporate glutamine in their proteins? Choose the correct answer.

(a) Glutamine is not present in the newly synthesized bacterial protein. Post translational modification converts glutamate to glutamine at the required sites.

(b) In these bacteria, the aminoacyl tRNA synthetase specific for tRNA glutamate (tRNA^Glu) also charges tRNA^Glnwith glutamine.

(c) In these bacteria, the aminoacyl tRNA synthetase specific for tRNA^Glu also charges tRNA^Gln with glutamate. A second enzyme then converts the glutamate of the charged tRNA^Glnto glutamine.

(d) In these bacteria, the aminoacyl tRNA synthetase charges tRNA^Glu with either glumate or glutamine acccording to their requirement during protein synthesis.

Ans. (c)

Sol. In these bacteria, the aminoacyl tRNA synthetase specific for tRNA^Glu also charges tRNA^Gln with glutamate. A second enzyme then converts the glutamate of the charged tRNA^Glnto glutamine.

41. As topoisomerases play an important role during replication, a large number of anticancer drugs have been developed tht inhibit the acitivity of these enzymes. Which of the following statements is not true about topoisomerases as a potential anticancer drug target?

(a) As cancer cells are rapidly growing cells, they usually contain higher level of topoisomerases.

(b) The transient DNA breaks created by topoisomerases are usually converted to permanent breaks in the genome in the presence of topoisomerase drugs.

(c) As cancer cells often have impaired DNA repair pathways, they are more susceptible towards topoisomerase targeted drugs.

(d) The drugs which specifically target topoisomerases, usually do not affect normal fast growing cells.

Ans. (d)

Sol. Drugs can't differenciate between topo-isomerases in normal cells and cancerous cells. Hence, drugs inhibiting topo-isomerases will target both cells.

42. Transposons can be primarily categorized into two types, DNA transposons are retrotransposons. Given below is some information regarding the above.

P. Eukaryotic DNA transposons excise themselves from one place in the genome and integrate into another site.

Q. Retrotransposons are RNA sequences that are, first reverse transcribed into cDNA and then integrate into the genome.

R. Retrotransposons move by a copy and paste mechanism through an RNA intermediate.

S. As DNA transposons move via a cut and paste mechanism, there can never be an increase in the copy number of a transposon.

Which of the statement(s) is/are true?

(a) P and R only

(b) Q ans S only

(d) S only

Ans. (a)

Sol. Transposable elements (TEs) are DNA sequences that have the ability to move from one chromosomal location to another through their integration into the genome at different sites. TEs have been found in virtually all eukaryotic organisms, covering 3-80% of their genomes and considerably influencing evolutionary history.

TEs can be grouped in two major classes. Class II elements or DNA transposons account for roughly 3% of the genome and mvoe by a cut and paste mechanism through an element encoded transposase. There is no indication of their activity to date whic h is believed to have subsided 37 million years ago. Class I elements or retrotransposons move by a copy and paste mechanism through an RNA intermediate which is reverse transcribed and then inserted at a new site in the host genome. The process is known as retrotransposition. Retrotransposons can be further categorized in two major subclasses : long terminal repeat retrotransposons (LTR) and non-LTR retrotransposons. LTR retroelements include endogenous retroviruses (ERVs) and consist 8% of the human genome.

43. Some errors occur during DNA replication that are not corrected by proofreading activity of DNA poymerase. These are corrected by specialized repair pathways. Defect in the activities of some of the following enzymes impair this process.

P. DNA polymerase III and DNA ligase

Q. AP endonuclease and DNA glycosidase

R. Mut S and Mut L

S. Rec A and Rec F

Defect in which of the above enzymes impair the process?

(a) P, Q and R

(b) S and Q

(d) P and R

Ans. (d)

Sol. Impaired repair mechanism mentioning in question is mismatch repair mechanism and protein involves in this mechanism are DNA polymerase III, DNA ligase, Mut S and Mut L.

44. An eukaryotic cell undergoing mRNA synthesis and processing was incubated with ³²P labelled ATP, with the label at the -position. Where do you think the radioactive isotope will appear in the mature mRNA?

(a) ³²P will not appear in the mature mRNA under any circumstances because and -phosphates are released during transcription.

(b) Phosphate groups of the phosphodiester backbone of the mRNA will be uniformly labelled as only -phosphates are released during transcription.

(d) No ³²P will appear in the nature mRNA because the 5'-terminal phosphate of an "A" residue will be further removed during the capping process.

Ans. (c)

Sol. First nucleotide of mRNA have both alpha and beta phosphate even after 5' capping. So, ³²P will appear at the 5'-end of the mRNA if only it has "A" as the first nucleotide.

45. One of the cellular events that TOR, a kinase, positively regulates is the rate of rRNA synthesis. TOR regulates the association of a transcription factor to a Pol I subunit. When TOR is inhibited by the drug rapamycin, the transcription factor dissociates from Pol I. A yeast strain is engineered, which expresses a fusion of the transcriptoin factor and the Pol I subunit. The level of rRNA synthesis is monitored in these cells using pulse labelling following rapamycin addition for the times indicated below. The transcript profile of rRNA observed for the wild type cells is given below:

Identify the pattern expected in the engineered strain:

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Engineered east strain have fused transcription factors with polymease I. It will remain unaffected from drug rapamycin and will show good expressions at each time.

46. Entry of enveloped viruses into its host cells is mediated by

(a) Only endocytosis

(b) both endocytosis and phagocytosis

(d) only pinocytosis

Ans. (c)

Sol. Both viruses and synthetic vectors must overcome formidable cellular membranes to gain entry into the host cell. While certain enveloped viruses undergo fusion with the plasma membrane concomitant to receptor binding, most nonenveloped viruses and synthetic vectors as well as some types of enveloped viruses, as well as some types of enveloped viruses, undergo cellular internalization via endocytosis and must subsequently escape endocytic vesicles to avoid degradation in the endo-lysosomal pathway.

47. Following are some of the characteristics of MHC class I and class II molecules except one which is applicable only for MHC class I. Identify the appropriate statement.

(a) They are expressed constitutively on all nucleated cells.

(b) They are glycosylated polypeptides with domain structure.

(d) They are expressed on surface membrane of B cells.

Ans. (a)

Sol. MHC molecule experssion determines whether T cells will recognize foreign antigen thereby leading to an adaptive immune response. Given their importance in adaptive immunity, the MHC genes are tightly regulated. In contrast to class I MHC molecules, which are expressed on alomst all nucleated cells, constitutive expression of class II MHC molecules is confined to APCs, such as dendritic cells, macrophages, B lymphocytes and thymic epithelial cells. Nonetheless, class I and class II MHC molecule expression can be induced by immune regulators and on cell activation.

48. Which of the following bacteria has sub-cellular localization in lysosomes?

(a) Salmonella typho

(b) Streptococcus pneumonia

(d) Mycobacterium tuberculosis

Ans.

Sol. Mycobacterium tuberculosis is a facultative intracellular pathogen that has evolved the bility to survive and multiply within human macrophages. M. tuberculosis and M. leprae are considered to be prototypical intracellular pathogens that have evolved strategies to enable growth in the intracellular pathogens that have evolved strategies to enable growth in the intracellular phagosomes. Mycobacterium tuberculosis bacteria has subcellular localization in lysosomes.

49. Which one of the following best defines an oncogene?

(a) An oncogene never codes for a cell cycle protein, which promotes cell proliferation.

(b) Oncogenes are always involved in inherited forms of cancer.

(d) An oncogene is a dominantly expressed mutated gene that renders a cell advantageous towards survival.

Ans. (d)

Sol. An oncogene is a mutated form of a normal cellular gene (the proto-oncogene) which codes for a protein which is either controlled abnormally so that it is expressed in abnormally large amounts or has gained activity so that it is more active than the normal protein. In either case only one gene of a pair needs to becoem oncogenic for the activity to be expressed, so the mutation is dominant. Some oncogenes code for a protein which forms part of a signal transduction pathway. A gain of function mutation in such a gene can make the pathway more active. Other oncogenes code for a protein which protects the cell against apoptosis. Increased production of this protein can prevent the cell entering apoptosis. However, either of these examples does not give a full definition of an oncogene. Cell cycle control proteins prevent cell division and are anti-oncogenic.

50. Which one of the following statements about receptor - enzyme is false?

(a) A receptor - enzyme has an extracellular ligand binding domain, a transmembrane domain and an intracellular catalytic (enzyme) domain.

(b) Many types of receptor enzymes are found in animals.

(d) Receptor - enzymes interact directly with intracellular G-proteins.

Ans. (d)

Sol. Activation of enzyme-linked receptors enables binding and activation of many intracellular signalling proteins, leading to changes in gene transcription and in many cellular functions. There are five families of enzyme-linked transmembrane receptors: Receptor tyrosine kinase (RTK) family. In summary, enzyme-linked receptors essentially turn an extracellular chemical signal into enzyme activity inside the cell. Specifically the most well-known of those are receptor tyrosine kinases. These are the largest and most well-known group.

51. Influenza virus (IV), a well known enveloped animal virus, enters its host cells through membrane fusion process catalyzed by hemagglutinin (HA) protein inside endosomes at 37°C. HA is localized in the lipid bilayer membrane of the IV as an integral membrane protein and is responsible for binding and fusion of IV membrane with the endosomal membrane of host cells. Upon binding, IV is internalized into cells through receptor mediated endocytosis followed by fusion of the IV membrane with endosome membrane catalyzed by HA. In a situation, if we wish to fuse IV membrane with its host cells (deficient in endocytosis) at the plasma membrane, mention the correct condition out of the following:

(a) Pre-treat IV in pH 5.0 followed by its binding and fusion with host cells at pH 7.4 and 37°C.

(b) Allow the IV to bind and fuse with host cells at pH 7.4 and 37°C.

(d) IV is subjected to incubation at 60°C for 30 minutes and allowed to bind and fuse with host cells, at pH 5.0 and 37°C.

Ans. (c)

Sol. Those that are activated at less acidic pH (~6.0) are thought to mediate fusion with early endosomes, whereas those with a lower pH threshold (~5.0) appear to direct the virus entry from late endosomes.

52. Glycophorin of red blood cell (RBC) membrane spans the membrane only once and the N-terminal is projected extracellularly and the C-terminal is exposed to the cytosolic side with the help of antibodies (labelled with fluorophores) against N-terminal and C-terminal peptides, orientation of glycophorin across membrane can be verified. Which one of the following statements is correct?

(a) Intact RBC can be labelled with C-terminal antibody.

(b) Permeabilized RBC can be labelled with C-terminal antibodies as well as N-terminal antibodies.

(d) Inside out ghost of RBC can be labeled with N-terminal antibodies.

Ans. (b)

Sol. Glycophorins are heavily glycosylated sialoglycoproteins of human and animal erythrocytes. Glycophorins play an important role in the invasion of red blood cells (RBCs) by malaria parasites, which involves several ligands binding to RBC receptors. Permeabilized RBC can be labelled with C-terminal antibodies as well as N-terminal antibodies.

53. Immunoglobulins have therapeutic applications in cancer treatment, infection clearance and targeted drug delivery. For this reason, immunoglobulins are briefly cleaved by the enzyme pepsin. Following are some of the statements regarding the brief digestion of immunoglobulin by pepsin.

P. F(ab)₂ fragment is generated which retains the antigen binding activity.

Q. F(ab) fragment having antigen binding activity and the crystallisable Fc fragment are generated.

R. The fragment generated on incubation with a proper antigen forms a visible precipitate.

S. The fragment generated is incapable of forming a visible precipitate on incubation with a proper antigen.

Which of the above statements are correct?

(a) P and Q

(b) P and R

(d) Q and R

Ans. (b)

Sol. The F(ab) fragment is an antibody structure that still binds to antigens but is monovalent with no Fc portion. An antibody digested by the enzyme papain yields two F(ab) fragments of about 50 kDa each and an Fc fragment. A visible antigen-antibody complex is called a precipitin, and in vitro assays that produce a precipitin are called precipitin reactions. A precipitin reaction typically involves adding soluble antigens to a test tube containing a solution of antibodies. Each antibody has two arms, each of which can bind to an epitope.

54. In an experiment peritoneal macrophages were isolated from strain A of guinea pig. These cells were sthen incubated with an antigen. After the antigen pulsed macrophages processed the antigen and presented it on their surface, these were mixed with T cells from (i) strain A or (ii) strain B (a different strain of guinea pig) or (iii) F1 progeny of strain A × B. T cell proliferation was measured in response to antigen pulsed macrophages. T-cells of which strain of guinea pig will be activated?

(a) Strain A only

(b) Strain B only

(d) Strain B and F1 progeny

Ans. (c)

Sol. The expansion of a T cell population by cell division. Follows T cell activation."T cells expressing higher affinity antigen receptors are likely to participate in spontaneous proliferation, although this hypothesis needs to be tested. The proliferating cells turn into memory phenotype cells and play a central role in regulating spontaneous proliferation of naive T cells.

55. Cadherins mediate Ca²⁺-dependent cell-cell adhesion and play an important role in embryonic development by changing the adhesive properties of cell. Aggergation of nerve cells to form an epithelium is correlated with the appearance of N-cadherins on cell surface and vice versa. N-CAM (neural cell adhesion molecules) belongs to Ig SF (immunoglobulin super family) and invovled in fine tunign of adhesive interaction. In order to see the effect of mutations of N-cadherin and and N-CAM. Which of the following results is most likely to occur?

(a) Mice of both set A and set B will die in early development.

(b) Mice of set A will die in early development but mice of set B will develop normally and show mild abnormalities in the development of nervous system.

(c) Mice of set A will show mild abnormalities in the development of nervous system whereas mice of set B will die early in development.

(d) Mice of set A and set B develop normally as other cell adhesion will compensate for the mutations.

Ans. (b)

Sol. N-Cadherin belongs to a superfamily of calcium-dependent transmembrane adhesion proteins. It mediates adhesion in the intercalated discs at the termini of cardiomyocytes thereby serving as anchor for myofibrils at cell-cell contacts. N-cadherin in the neural plate is required for the morphogenic movements of neurulation. In most vertebrates, including humans, mice, frogs and chickens, the formation of the central nervous system starts with the invagination of the flat sheet of epithelial cells of the neural plate into a hollow tube. The neural cell adhesion molecule (NCAM) is an immunoglobulin-like neuronal surface glycoprotein which binds to a variety of other cell adhesion proteins to mediate adhesion, guidance, and differentiation during neuronal growth. Mutations in the L1 neural cell adhesion molecule, a transmembrane glycoprotein, cause a spectrum of congenital neurological syndromes, ranging from hydrocephalus to mental retardation.

56. Virus infects a particular cell type, integrates its genome into a site that contains a proto-oncogene, transforms the cell and increases the evel of protein 'X', which increases cellular proliferation. A compound 'P' is known to increase the level of tumor suppressor proteins in that cell type whereas a compound 'Q' helps in stimulating a protein Z that can bind to 'X' rendering it inactive. Which one of the following graph correctly represents the mode of action of 'P' and 'O'?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Tumor suppressor genes represent the opposite side of cell growth control, normally acting to inhibit cell proliferation and tumor development. In many tumors, these genes are lost or inactivated, thereby removing negative regulators of cell proliferation and contributing to the abnormal proliferation of tumor cells. Proto-oncogenes encode proteins that function to stimulate cell division, inhibit cell differentiation, and halt cell death. All of these processes are important for normal human development and for the maintenance of tissues and organs. In graph C, when neither of the gene is present, will lead to increased cell proliferation. When either of the gene is present will cause decreased cell proliferation comparatively and when both of them are present will cause more less cell proliferation.

57. Cancer is often believed to arise stem cells rather than fully differentiated cells. Following are certain views related to the above statement. Which one of the following is not correct?

(a) Stem cells do not divide and therefore require fewer changes to become a cancer cell.

(b) Cancer stem cells can self-renew as well as generate the non-stem cell populations of the tumor.

(d) Stemness genes can often function as oncogenes.

Ans. (a)

Sol. Stem cells survive much longer than ordinary cells, increasing the chance that they might accumulate genetic mutations. It might take only a few mutations for one cell to lose control over its self-renewal and growth and become the source of cancer.

58. Following are certain statements that describe plant-pathogen interactions:

P. Hemibiotrophic pathogens are characteized by initially keeping host cells alive followed by extensive tissue damage during the later part of the infection.

Q. Effectors are molecules present in host plants that act against the pathogen attack.

R. Plants possess pattern recognition receptors (PRRs) that perceive microbe-associated molecular patterns (MAMPs) present in specific class of microorganisms but are absent in the hosts.

S. Phytoalexin production is a common mechanism of resistance to pathogenic microbes in a wide range of plants.

Which one of the following combinations is correct?

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (b)

Sol. An organism is hemibiotrophic pathogen that is parasitic in living tissue for some time and then continues to live in dead tissue. MAMPs are recognized by pattern recognition receptors (PRRs), which are localized on the surface of plant cells; this first phase of defense induction is called MAMP-triggered immunity (MTI). Phytoalexins are low molecular weight antimicrobial compounds that are produced by plants as a response to biotic and abiotic stresses. As such they take part in an intricate defense system which enables plants to control invading microorganisms.

59. Bones of vertebrates are derived from embryonic

(a) ectoderm

(b) epiderm

(d) endoderm

Ans. (c)

Sol. Cranial and Cardiac Mesoderm

In the head region, unlike other regions of the embryo where bone is derived exclusively from mesoderm, the bones of the cranium, face and neck are formed from ectoderm-derived neural crest cells as well as msoderm. The muscles of the face an neck, with the exception of the iris muscles are derived from the anterior paraxial mesoderm.

60. During development, if a cell has committed to a particualr fate, it is said to be

(a) pluripotent

(b) totipotent

(d) differentiated

Ans. (c)

Sol. Vertebrate cells start to become different after the eight cell stage as a result of cell-cell interactions like those we just will influence the future developmetn fate of the cells is determined. The commitment of a particular cell to a specialized developmental path is called determination. A cell in the prospective brain region of an amphibian embryo at the early gastrula stage has not yet been determined; if transplanted elsewhere in the embryo, it will deelop like its new neighbours. By the late gastrula stage, however, determination has taken place and the cell will develop as neural tissue no matter where it is transplanted.

61. The initial dorsal-ventral axis in amphibian embryos is determined by

(a) the point of sperm entry

(b) gravity

(d) genetic differernces in the cells

Ans. (a)

Sol. It embryos without localized factors or where localized factors establish some but not all axes subsequent interactions between the early embryo and external cues are thought to "break the symmetry" of the egg. For instance, the sperm entry point in amphibian embryos or the site where mammalian embryos first contact the uterine wall may provide cues that polarize the formation of additional axes in the embryo. In the cse of amphibian embryos the sperm entry point establishes the plane of left-right symmetry and the site of the primary lip where the anterior most cell identities are established and gastrulation begins.

62. Sperm cell behaviour during double fertilization in Arabidopsis can be stated as follows. Identify the incorrect statement:

(a) Polen tube bursts and discharges sperm cells.

(b) Sperm cells produce pollen tubes and enter into female gametophyte.

(d) One sperm nucleus fuses with the egg cell and the other fuses with the central cells.

Ans. (b, c)

Sol. Antipodal cells leads to the formation of the typical female gametophyte or embryo sac Synergids and antipodals degenerates after fertilization.

63. Which one of the following combinations is the correct pairing of ligands with their receptors?

P. FGF 1. Patched

Q. Hedgehog 2. Frizzled

R. Wnt 3. Receptor tyrosine kinase

(a) P-3, Q-1, R-2

(b) P-1, Q-3, R-2

(d) P-3, Q-2, R-1

Ans. (a)

Sol.

64. Given are certain facts which define 'determination' of a developing embryo.

P. Cells have made a commitment to a differentiation program.

Q. A phase where specific biochemical actions occur in embryonic cells.

R. The cell cannot respond to differentiation signals.

S. A phase where inductive signals trigger cell differentiation.

Which of the above statements best define determination?

(a) Q and S

(b) P and R

(d) Only Q

Ans. (c)

Sol. Cell determination, in which initially identical cells become committed to different pathways of development. A fundamental part of cell determination is the ability of cells to detect different chemicals within different regions of the embryo.

65. What would happen as a result of a transplanation experiment in a chick embryo where the leg mesenchyme is placed directly beneath the wing apical ectodermal ridge (AER)?

(a) Distal hindlimb structtures develop at the end of the limb.

(b) A complete hindlimb will form in the region where the forelimb should be.

(d) Neither a forelimb for a hindlimb would form since the cells are already determined.

Ans. (a)

Sol. If leg mesenchyme is placed directly beneath the wing AER, distal hindlimb structures (toes) develop at the end of the limb. (However, if this mesenchyme is placed farther from the AER, the hindlimb mesenchyme becomes integrated into wing structures).

66. If you remove a set of cells from an early embryo, you observe that the adult organism lacks the structure that would have been produced from those cells. Therefore, the organism seems to have undergone

(a) autonomous specification

(b) conditional specification

(d) syncytial specification

Ans. (a)

Sol. Autonomous Specification : In this case, if a particular blastomere is removed from an embryo early in its development, that isolated blastomere will produce the same cells that it would have made if it were still part of the embryo. Moreover, the embryo from which that cell is taken will lack those cells (and only those cells) that would have been produced by the missing blastomere. Autonomous specification gives rise to a pattern of development referred to as mosaic development, since the embryo appears to be constructed like a tile mosaic of independent self-differentiating parts. Invertebrates embryos, especially those of molluscs, annelids and tunicates, often use autonomous specification to determine the fate of their cells.

67. Dose-dependence of retinoic acid treatment supports the notion that a gradient of retinoic acid can act as a morphogen along the proximo-distal axis in a developing limb. Following are certain facts related to the above nation.

P. Treatment with high level of retinoic acid causes a proximall blastema to be respectfied as a distal blastema and only distal structures are regenerated.

Q. Treatment with high level of retinoic acid causes a distal blastema to be erspecified as a proximal blastema and regeneration of a full limb may be initiated.

R. Treatment with retinoic acid affects only distal blastemas and causes them to form only proimal structures.

S. Treatment with high level of retinoic acid causes any blastema to form only distal structures.

Which one of the following is correct?

(a) Q and S

(b) Only R

(d) Only Q

Ans. (d)

Sol. Retinoic acid (RA) is a morphogen derived from retinol (vitamin A) that plays important roles in cell growth, differentiation, and organogenesis. Treatment with high levels of retinoic acid causes a distal blastema to be respecified as a proximal blastema, and regeneration of a full limb may be initiated from proximal values.

68. Match the two columns following asexual reproduction of plants and apomixes:

P. Agamospermy 1. No seed information

Q. Clonal propagation 2. Seed formation

R. Embryo sac formed from nucellus 3. Diplospory

or integument of the ovule

S. Gametophyte develops without 4. Apospory

fertilization from unreduced megaspore

(a) P-1, Q-2, R-3, S-4

(b) P-2, Q-3, R-1, S-1

(d) P-2, Q-1, R-4, S-3

Ans. (d)

Sol. Agamospermy is a type of asexual reproduction in which plant propagation occurs by means of apomictic seeds (or asexual seeds). In agamospermy, one or more sporophytic cells undergo stages of embryogeny and eventually grow into the mature embryo. Multiplication of genetically identical copies of a cultivar by asexual reproduction is called clonal propagation and a plant population derived from a single individual by asexual reproduction constitutes a clone. Diplospory is the method by which the megaspore mother cells divide by mitosis thrice to give rise to 8 progeny plants. In this process no sexual reproduction or mixing of genes are involved. This process is similar to asexual reproduction. Apospory is the development of 2n gametophytes, without meiosis and spores, from vegetative, or nonreproductive, cells of the sporophyte.

69. According to the ABC model of floral development in Aabidopsis as shown below, several genes/transcription factors e.g. AP1, AP2, AP3, AG etc. are involved.

Which one of the following statement is correct?

(a) Apetala 2 (AP2) transcripts experssed during sepal and petal development.

(b) Agamous AG is considered as class A gene.

(d) AP3 expressed during sepal development.

Ans. (a)

Sol. AP2 transcripts are ubiquitous throughout the developing flower, AP2 function is limited to the first and second whorls i.e, sepal and petal development.

70. Rhizobial genes that participates in legume nodule formation are called nodulation (nod) genes. The nodD-encoded protein

(a) is an acetyl transferase that adds a fatty acyl chain to the Nod factor.

(b) binds to the nod box and induces transcription of all nod genes.

(d) influences the host specificity of Rhizobium.

Ans. (b)

Sol. Nod-D is constitutively expressed in the rhizobia. Once activated, the rhizobial Nod-D protein induces the transcription of the other nod genes by binding to the highly conserved nod box present in the promoter region of all nod genes except nod-D. These nod genes code for nodulation proteins that are involved in the synthesis of Nod factors, that is, signaling molecules released by rhizobia that induce infection with rhizobia and nodule organogenesis.

71. Which one of the following plant hormones uses the two component histidine kinase receptor system for signal transduction?

(a) Auxin

(b) Gibberellin

(d) Abscisic acid

Ans. (c)

Sol. The Cytokinin Receptor : The first indication that two component type proteins might be involved in cytokinin signaling came from the identification of the Arabidopsis histidine kinase encoding gene, CKI1, as a gene that conferred cytokinin independent (CKI) growth to callus cells when over expressed. A second sensor histidine kinase CRE1 was subsequently isolated in an EMS-mutagenesis screen for callus tissue that displayed reduced shoot formation and greening in response to exogenous cytokinin.

72. Which one of the following photorreceptors plays a role in day length perception and circadian rhythms?

(a) Zeitlupe family

(b) Cryptochromes

(d) UV resistance locus 8

Ans. (a)

Sol. Zeitlupe is a circadian photoreceptor stabilized by Gigantea in blue light. The Zeitlupe (ztl) gene family has emerged as a novel set of proteins that may fuse photoreception with protein degradation.

The founding member of the ztl gene family was first identified in a screen for period length mutants in Arabidopsin. Ztl-1 has an endogenous free-running period of 27 h period, 3 h longer than wild type. All circadian ouptpus thus far tested are lengthened by this mutation, though some rhythms are more strongly affected than others.

73. Which one of the following is the correct order of electron transport during light reaction in the thylakoid membrane of chlooplast?

(a) P680 Cytochrome b₆f PC PQ

(b) P680 PC Cytochrome b₆f PQ

(b) P680 PQ Cytochrome b₆f PC

Ans. (d)

Sol.

Figure : Electron transport during light reaction in the thylakoid membrane of chloroplast.

74. In a mitochondrial respiration experiment, a researcher observed the following profile of oxygen consumption upon addition of following compounds at times P, Q and R.

1. ADP + P_i

2. Dinitrophenol, an uncoupler

3. Oligomycin, an ATPase inhibitor

4. Cyanide

5. Succinate

Which of the following describes the profile appropriately?

(a) P-2, Q-4, R-5

(b) P-1, Q-4, R-3

(d) P-1, Q-13, R-2

Ans. (d)

Sol. Oligomycin inhibits respiration, but in contrast to electron transport inhibitors, it is not a direct inhibitor of the electron transport system. Instead, it inhibits the proton channel of ATP synthase. Mitochondrial uncouplers, such as 2,4 dinitrophenol (DNP), increase the cellular respiration by decreasing mitochondrial membrane potential (delta psi). ADP can be converted, or powered back to ATP through the process of releasing the chemical energy available in food.

75. Indicate the names of the following molecules:

(a) P = isocitrate, Q = -ketoglutarate, R = oxaloacetate, S = citrate

(b) P = citrate, Q = isocitrate, R = -ketoglutarate, S = oxaloacetate

(d) P = citrate, Q = isocitrate, R = oxaloacetate, S = -ketoglutarate

Ans. (b)

Sol. In order to experimentally examine the evolutionary potential latent in the hydrothermal circulation, we ran the flow reactor for the reaction solution comprising all of the eight major kinds of carboxylic acid molecules constituting the citric acid cycle, that is oxaloacetate, citrate acid cycle that is, oxaloacetate, citrate, isocitrate, alpha-ketoglutarate, succinate, fumarate, L-malate and pyruvate serving as both the energy and carbon sources to the cycle.

76. Constitutive photomorphogenesis (COP1) protein, an E3 ubiquitin ligase, regulates the turnover of proteins required for photomorphogenic development. Following are certain independent statements related to the funtion of COP1 protein:

P. In light, COP1 along with SPA1 adds ubiquitin tags to a subset of nuclear proteins.

Q. The proteins ubiquitinated by COP1 and SAP1 are targeted for degradation by the 26S proteasome.

R. In dark COP1 is slowly exported to the cytoso from nucleus.

S. The absence of in the nucleus permits the accumulation of transcriptional activators necessary for photomorphogenic development.

Which one of the following combinations is correct?

(a) P and R

(b) P and S

(d) Q and S

Ans. (d)

Sol. Arabidopsis COP1 is a constitutive repressor of photomorphogenesis that interacts with photomorphogenesis-promoting factors such as HY5 to promote their proteasome-mediated degradation. SPA1 is a repressor of phytochrome A-mediated responses to far-red light. In the dark, COP1 accumulates in the nucleus, mediating HFR1 degradation. In light, COP1 translocates to the cytosol, which allows the accumulation of HFR1 to promote photomorphogenesis

77. The following statements are made to describe auxin signal transduction pathway, from receptor binding to the physiological response:

P. Auxin response factor (ARFs) are nuclear proteins that bind to auxin response elements (Aux REs) to activate or repress gene transcription.

Q. AUX/IAA proteins are secondary regulators of auxin-induced gene expression. Binding of AUX/IAA proteins to the ARF protein blocks its transcription regulation.

R. Auxin binding to TIRI/AFB promotes ubiquitin-mediated degradation and removal of AUX/IAA proteins.

S. Auxin binding to auxin response factors (ARFs) causes their destruction by the 26S proteasome pathway.

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

(a) P, Q and R

(b) P, R and S

(d) P, Q and S

Ans. (a)

Sol. Auxin response factors (ARFs) are transcription factors that bind to TGTCTC auxin response elements in promoters of early auxin response genes. Aux/IAA proteins are short-lived nuclear proteins comprising several highly conserved domains that are encoded by the auxin early response gene family. These proteins have specific domains that interact with ARFs and inhibit the transcription of genes activated by ARFs. Auxin (pink) binds to TIR1 (blue) or a TIR1-Aux/IAA complex to promote or stabilize TIR1-Aux/IAA domain II interaction. TIR1 tethers the Aux/IAA protein to an SCF complex (purple) that is thought to catalyze attachment of multiple ubiquitin (Ub) moieties to the Aux/IAA target protein.

78. Light reactions of photosynthesis are carried out by four major protein complexes: Photosystem I (PSI), photosystem II (PSII), the cytochrome b₆f complex and ATP synthase. The following are certain statements on PSI:

P. PSI reaction centre and PSII reaction centre are uniformly distributed in the granallamellae and stromal lamellae.

Q. The electron donor for the P700 of PSI is plastocyanin and electron acceptor P700* is a chlorophyl known as A₀.

R. The core antenna and P700 are bound to two key proteins PsaA and PsaB.

S. Cyclic electron flow occurs from the reducing side of PSI via plastohydroquinone and b₆f compex. This supports ATP synthesis but does not reduce NADP⁺.

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

(a) P, Q and R

(b) P, R and S

(d) Q, R and S

Ans. (d)

Sol. Cyclic Phosphorylation :

Non-Cyclic Phosphorylation :

79. Ribulose bisphosphate carboxylase (Rubisco) catalyzes both carboxylation and oxygenation of ribulose-1, 5-bisphosphate. The latter reaction initiates a physiological process known as 'photorespiration'. The following are certain statements on photorespiration.

P. The active sites on Rubisco for carboxylationand oxygeneation are different.

Q. One of the steps in photorespiration is conversion of glycine to serine.

R. 50% of carbon lost in chlorophlast due to oxygen-ation is recovered through photorespiration.

S. The pathway of photorespiration invoves chloroplast, peroxisome and mitochondria.

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

(a) P and R

(b) P and S

(d) R and S

Ans. (c)

Sol. Photorespiration involves three organelles (chloroplasts, peroxisomes, and mitochondria), each with unique transport mechanisms for the cycle's intermediates.

80. Several transport steps are involved in the movement of photosynthate from the chloroplasts. Following are certain statements regarding the transport of photosynthate:

P. Pentose phosphate formed by photosynthesis during the day is transported from the chloroplast to the cytosol, where it is converted to sucrose.

Q. Carbon stored as starch exits the chloroplast at night primarily in the form of maltose and is converted to sucrose in cytosol.

R. During short distance transport, sucrose moves from producing cells in the mesophyl to cells in the vicinity of the sieve elements in the smallest veins of the leaf.

S. In the process of pholem loading, sugars are transported into phloem parenchyma cells.

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

(a) P and Q

(b) Q and R

(d) P and S

Ans. (b)

Sol. The results presented here indicate that carbon from starch breakdown at night is normally exported from chloroplasts as hexose ( maltose) and converted to sucrose in cytosol. At the short-distance level within organs, sucrose migrates from mesophyll to phloem via the symplast and apoplast. At the long-distance level between organs, bulk flow within sieve tubes transports phloem sap from sugar sources to sugar sinks.

81. Match the following associations involved in dinitrogen fixation with their representative genera

Associations Genera

P. Heterotrophic nodulate 1. Azotobacter

Q. Heterotrophic nonnodulate 2. Frankia

R. Phototrophic associative 3. Nostoc

S. Phototrophic free living 4. Rhodospirillum

(a) P-2, Q-1, R-4, S-3

(b) P-3, Q-1, R-2, S-4

(d) P-2, Q-1, R-3, S-4

Ans. (a, d)

Sol. Table : Organisms and associations involved in dinitrogen fixation

*Nitrogen-fixing microbes are heterotrophic bacteria, if they get their organic carbon from the environment. They are phototrophic bluegreen algae, if they produce it themselves through photosynthesis. Some forms of both microbial groups are typically associated with plants, whereas others are free living.

82. Insulin increases facilitated diffusion of glucose in muscle cells by

(a) phosphorylation of glucose transporters.

(b) translocation of glucose transporter containing endosomes into the cell membrane.

(d) dephosphorylation of glucose transporters.

Ans. (b)

Sol. Glucose is an important fuel for contracting muscle and normal glucose metabolism is vital for health. Glucose enters the muscel cell via facilitated diffusion through the GLUT4 glucose transporter which translocates from intra-cellular storage depots to the plasma membrane and T-tubules upon muscle contraction.

83. The transport of fructose into the enterocytes is mediated by

(a) sodium-dependent glucose transporter 1 (SGLT 1)

(b) glucose transporter 5 (GLUT5)

(d) GLUT 4

Ans. (b)

Sol. GLUT5 is an apical membrane transporter that transports fructose, a ketohexose into the enterocyte. Unlike GLUT2, GLUT5 is a low affinity high capacity transporter with a K_m for fructose measured in Xenopus oocytes to be 5 to 6 mmol/L. Some GLUT5 transporters are also localized in the basolateral membrane, where they may complement the GLUT2-mediated exit of fructose the enterocyte.

84. The cell bodies of sympathetic-preganglionic neurons are located in

(a) intermediolateral cell column of spinal cord

(b) posterior cell column of spinal cord

(d) paravertebral ganglion

Ans. (a)

Sol. Sympathetic preganglionic neurons are small-to medium-sized cholinergic neurons and their cell bodies are located in the thoracic and upper lumbar spinal cord in four distinct subnuclei within the lateral horn. The majority of sympathetic preganglionic somata occur in the intermediolateral cell column, which lies at the border between the grey and white matter of the spinal cord. In the intermediolateral cell column, the cell bodies of sympathetic preganglionic neurons are spindle-shaped or fusiform and occur in groups or "nests" that are spaced at short intervals along the grey-white boundary. Most of the dendrites of the neurons in th intermed iolateral cell column run rostrally or caudally and can be hundreds of micrometers long. In addition, some sympathetic preganglionic neurons in the intermediolateral cell column have dendrites that are oriented into the dorsolateral funciulus or toward the central canal.

85. The di- and tripeptides are transported in the enterocytes by peptide transporter 1 that requires

(a) Na⁺

(b) Ca⁺⁺

(d) Cl^–

Ans. (c)

Sol. For the absorption of di and tri peptides, only one transport system, designated as PepT1 is known. This is a low affinity, high capacity transport system and handles essentially all possible protein derived di and tri peptides as well as various peptidomimetics by PEPT1 which requires H⁺ ions.

(a) Active transport via PepT1 tarnsporter.

(b) Transcellular movement of CPP with peptides as cargo.

Figure : Potential routes of peptide uptake in enterocytes. (A) The primary route of di and tri peptide absorption is through contransport with H⁺ by the peptide transporter. PepT1 (B) Cell penetrating peptides (CPP) are capable of carrying cargo such as peptides to the inside of cells. (C) Increased permeability of tight junctions permits uptake of peptides via the paracellular route.

86. A majority of humans with normal colour vision was found to be more sensitive to red light in Rayleigh match where the subject mixed variable amount of red and green light to match monochromatic orange. Which one of the following statements is not true to explain the observation?

(a) There are variations in the sensitivity of long-wave cone pigments.

(b) The short-wave cone opsin in red-sensitive subjects is different from others.

(c) The absorption curve of long-wave cone pigment peaks at 556 nm in red-sensitive subjects while it peaks at 552 nm in others.

(d) The long-wave cone opsin in red-sensitive subjects is different in primary structure from that of others.

Ans. (b)

Sol. A majority of humans with normal colour vision was found to be more sensitive to red light in Rayleigh match where the subject mixed variable amount of red and green light to match monochromatic orange. Cone opsin in red sensitive subject is categorised as long wave cone opsin, green opsin as intermediate wave cone opsin and blue opsin as short wave cone opsin.

87. The membrane potential in a giant squid axon recorded intracellularly at the resting condition (–70 mV) was reversed at the peak of action potential (+35 mV) after stimulation of the nerve fibre with a threshold electrical stimulus. The overshoot of the membrane potential has been explained in the following proposed statements.

P. The rapid increase in Na⁺-conductance during early phase of action potential uses membrane potential to move toward the equilibrium potential of Na⁺ (+45 mV).

Q. The Na⁺-conductance quickly decreases toward resting level after peak in the early phase and Na⁺-ions are not able to attain its equilibrium potential within this short time.

R. The conductance of K⁺at the early phase of action potential is increased and that leads to the reversal of membrane potential.

S. The conductance of K⁺-conductance due to stimulation of nerve occurs before the changes of Na⁺ is initiated and thus causes overshoot at the peak of action potential.

Which one of the following is correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (b)

Sol. The overshoot of membrane potential occurs when .The rapid increase in Na⁺-conductance during early phase of action potential uses membrane potential to move toward the equilibrium potential of Na⁺ (+45 mV) and The Na⁺-conductance quickly decreases toward resting level after peak in the early phase and Na⁺-ions are not able to attain its equilibrium potential within this short time.

88. A person showed the symptoms of diarrhea, gas and pain whenever milk was consumed. The doctor advised the person to take curd instead of milk and subsequently the symptoms mostly disappeared due to this change of dairy product. The following statements are proposed to explain this observation:

P. The person has deficiency in the intestinal sucrase-maltase.

Q. Curd is not deficient in sucrose and maltose.

R. The person has deficiency in the intestinal lactase.

S. The bacteria in curd contain lactase.

Which one of the following is correct?

(a) P only

(b) P and Q

(d) R and S

Ans. (d)

Sol. A person showed the symptoms of diarrhea, gas and pain whenever milk was consumed. The doctor advised the person to take curd instead of milk and subsequently the symptoms mostly disappeared due to this change of dairy product. It occurs becaue, The person has deficiency in the intestinal lactase. and The bacteria in curd contain lactase.

89. A diabetic patient has a high blood glucose level due to reduced entry of glucose into various peripheral tissues in addition to other causes. There is no problem of glucose absorption, however, in the small intestine of these patients. The following statements are put forward to explain this observation:

P. Glucose is transported into the cells of muscles by glucose transporters (GLUTs) which are influenced by insulin receptor activation.

Q. Glucose transported into the enterocytes is mediated by sodium-dependent glucose transporters (SGLTs) which are not dependent on insulin.

R. Glucose molecules are transported in the small intestine by facilitated diffusion.

S. The secondary active transport of glucose occurs in muscles.

Which one of the following is incorrect?

(a) P only

(b) P and Q

(d) R and S

Ans. (d)

Sol. A diabetic patient has a high blood glucose level due to reduced entry of glucose into various peripheral tissues in addition to other causes. There is no problem of glucose absorption, however, in the small intestine of these patients. Glucose is transported into the cells of muscles by glucose transporters (GLUTs) which are influenced by insulin receptor activation. and Glucose transported into the enterocytes is mediated by sodium-dependent glucose transporters (SGLTs) which are not dependent on insulin.

90. Action potentials were recorded intracelularly from different parts of mammaian heart and these are shown below. Which one of these has been recorded from sinoatrial node?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Option (b) correctly represent action potential from different part of mammalian heart.

91. Which one of the following options correctly relates the source gland/organ with its respective hormone as well as function?

Source gland Hormone Function

(a) Thyroid Thyroxine Regulates blood calcium level

(b) Anterior pituitary Oxytocin Contraction of uterine muscles

nephron

(d) Corpus luteum Estrogen Supports pregnancy

Ans. (c)

Sol. Water Reabsorption : To distal nephron, 35% of filtered water goes (65% is absorbed in proximal cenvoluted tubule). The amount of water absorbed depends on need of the body for water and is regulated by hormone ADH (Antidiuretic hormone) from posterior pituitary gland. Antidiuretic hormone increases the absorption of water from distal nephron. If no ADH is secreted , no water is absorbed from distal nephron which leads to large amont of water excretion causing formation of dilute urine.

92. Which one of the following statement is incorrect?

(a) Quantitative inheritance results in a range of measurable phenotypes for a polygenic trait.

(b) Polygenic traits often demonstrate continuous variation.

(d) Alleles governing quantitative traits do not segregate and assort independently.

Ans. (d)

Sol. Alleles govering quantitative traits are from different genes. Hence, there locus may be distributed throughout genome. Hence, they may show independent assortment and segregation.

93. A mouse carrying two alleles of insulin-like growth factor II (IgF2) is normal in size; whereas a mouse that carries two mutant alleles lacking the growth factor is dwarf. The size of a heterozygous mouse carrying one normal and one mutant allele depends on the parental origin of the widl type allele. Such pattern of inheritance is known as one mutant allele depends on the parental origin of the wild type allele. Such pattern of inheritance is known as

(a) Sex-linked inheritance

(b) Genomic imprinting

(d) Cytoplasm inheritance

Ans. (b)

Sol. A mouse carrying two alleles of insulin-like growth factor II (IgF2) is normal in size; whereas a mouse that carries two mutant alleles lacking the growth factor is dwarf. The size of a heterozygous mouse carrying one normal and one mutant allele depends on the parental origin of the widl type allele. Such pattern of inheritance is known as one mutant allele depends on the parental origin of the wild type allele. Such pattern of inheritance is known as Genomic imprinting.

94. What is the genotype of a male Drosophila fly that has yellow body colour and red eyes. Brown (y⁺) is dominant over yellow (y) and red (w⁺) is dominant over white (w). Both are carried on X-chromosome.

(a) X^w+y Y

(b) X^wy Y

(d) X^wy+ X^wy+Y

Ans. (a)

Sol. Genotype of male drosophila with yellow body colored and red eyes would be X^w+y Y.

95. Poplar is a dioecious plant. A wild plant with 3 genes AABBCC was crossed with a triple recessive mutant aabbcc. The F1 male hybrid (AaBbCc) was then back crossed with the triple mutant and the phenotypes recorded are as follows:

AaBbCc 300

aaBbCc 100

aaBbcc 16

AabbCc 14

AaBbcc 65

aabbCc 75

aabbcc 310

Aabbcc 120

The distance in map unit (mu) between A to B and B to C is

(a) 25 and 17 mu, respectively

(b) 33 and 14 mu, respectively

(d) 33 and 17 mu, respectively

Ans. (a)

Sol.

96. A three point test cross was carried out in Drosophila melanogaster involving three adjacent X, Y and Z, arranged in the same order. The distance between X to Y is 32.5 map unit (mu) and that between X to Y is 20.5 map. The coefficient of coincidence = 0.886. What is the percentage of double recombinants in the progeny obtained from the testcross?

(a) ~6%

(b) ~8%

(d) ~16%

Ans. (a)

Sol. Typing error. The distance between Y to Z should be 20.5 map.

Expected double crossovers =

Coefficient of coincidence = ; given that, coefficeint of coincidence = 0.886

Observed double crossovers = co-efficient of coincidence × expected double crossovers

= 0.886 × 0.066625 0.06 × 100 = 6%

97. Two interacting genes (independently assorting) were involved in the same pathway. Absence of either genes function leads to absence of the end product of the pathway. A dihybrid cross involving the two genes is carried out. What fraction of the F2 progeny will show the presence of the end poduct?

(a) 1/4

(b) 3/4

(d) 15/16

Ans. (c)

Sol. As per question two interacting genes (independently assorting) were involved in the same pathway. Absence of either genes function leads to absence of the end product of the pathway. So, end product will be produce only in the F2 progeny having dominant form of both genes. Hence, probability will be 9/16.

98. A male mouse cell line has a large translocation from X-chromosome into chromosome 1. When a GFP containing transgene is inserted in this chromosome 1 with translocation, it is often silenced. However when inserted in the other homologue of chromosome 1 that does not contain the translocation, it is almost always expressed. Which of the following phenomenon best describes this effect?

(a) Genome imprinting

(b) Gene balance

(d) Dosage compensation

Ans. (d)

Sol. A male mouse cell line has a large translocation from X-chromosome into chromosome 1. When a GFP containing transgene is inserted in this chromosome 1 with translocation, it is often silenced. However when inserted in the other homologue of chromosome 1 that does not contain the translocation, it is almost always expressed. This affect is most probably dosage compensation.

99. Five bacterial markers were followed for a cotrans-duction experiment. The following table documents the observations of this experiment. '+' denotes co-transduction and '–' denotes lack of thereof; 'ND' stands for not determined. Pick the correct order in which the genes are arranged on the bacterial chromosomes.

(a) str – gal – leu – arg – met

(b) leu – met – arg – str – gal

(b) leu – str – met – gal – arg

(d) arg – gal – str – leu – met

Ans. (d)

Sol. The correct sequence of marker is arg – gal – str – leu – met.

100. Which of the following statements is not true regarding the closer affinity of Archaea to Eukarya than to Bacteria?

(a) Both Archaea and Eukarya lack peptidoglycan in their cell walls.

(b) The initiator amino acid for protein synthesis is methionine in both Archaea and Eukarya.

(d) In both Archaea and Eukarya the RNA polymerase is of several kinds.

Ans. (c)

Sol. Archea are similar to eukaryotes in the following respects.

1. The DNA of both archea and eukaryotes are associated with histone proteins. Bacterial DNA is not.

2. Ribosome activity in both archea and eukaryotes is not inhibited by the antibiotics streptomycin and chloramphenicol. In bacteria, ribosome activity is inhibited by these antibiotics.

101. Match the following larval forms with phyla that they occur in

Larva Phylum

P. Amphiblastula 1. Mollusca

Q. Nauplius 2. Echinodermata

R. Nauplius 3. Porifera

S. Glochidium 4. Arthropoda

5. Annelida

(a) P-3, Q-4, R-1, S-2

(b) P-4, Q-3, R-1, S-5

(d) P-5, Q-1, R-2, S-3

Ans. (a)

Sol. Amphiblastula- A larva of some sponges that has equal numbers of both flagellate and nonflagellate cells separated from each other on opposite sides of the blastula."Nauplius- a crustacean larva in usually the first stage after leaving the egg and with three pairs of appendages, a median eye, and little or no segmentation."The glochidium is a microscopic larval stage of some freshwater mussels." A bipinnaria is the first stage in the larval development of most starfish. The bipinnaria is free-living, swimming as part of the zooplankton

102. Which of the following National parks has the highest density of tigers among protected areas in the world?

(a) Jim Corbett

(b) Kaziranga

(d) Manas

Ans. (b)

Sol. Kaziranga has the highest density of tigers among the protected areas in the world and was declared as a tiger reserves in 2006. This park has large breeding population of elephants, water buffalo and swamp deer. Among the other animals, tiger, elephant, wild-boar, civet-cat, various deers and birds such as pelican, florician, stork and fishing eagle are important fauna of this park. Kaziranga is recognised as an important bird sancturary by the Birdlife International for Conservation of Avifaunal Species.

103. Which of the following is not a prediction arising out of Wilson-MacArthur's - Theory of Island Biogeography?

(a) The number of species on an island should increase with its size/area.

(b) The number of species should decrease with increasing distance of the island from the source pool.

(d) Species richness on an island should be related to its average distance to the neighbouring islands.

Ans. (c)

Sol. Island Biogeography Theory (MacArthur and Wilson, 1967) attempts to identify the processes which underlie equilibrium in species numbers on any island through studying the processes of immigration and extinction. The theory was inspired by three patterns which had long been recognized as pan of island ecology :

that larger islands tend to support more species.

that remote islands tend to support fewer species.

that there is often a turnover of species on island, with newcomers replacing other species that become extinct.

104. For a population growing exponentially with a growth rate r, its population doubling time is

(a) (N₀ × 2)r

(b) In 2/r

(d) In r × 2

Ans. (b)

Sol. By the doubling time we mean, as the name implies, the amount of time required for a given principal to double. The suprising fact here is that the doubling time does not depend on the principal P. To see why this is so in the continuous compounding case, we begin with the formula

A = Pe^rt

We are interested in the time t at which

A = 2P

2P = Pe^rt

2 = e^rt

rt = ln 2

t =

Denoting the doubling time by T₂, we have the following formula.

Doubling time = T₂ =

105. In which ecosystem is the autotroph-fixed energy likely to reach the primary carnivore level in the shortest time?

(a) Temperature deciduous forest

(b) Grassland

(d) Tropical rain forest

Ans.

Sol. In ocean ecosystem, autotroph-fixed energy is likely to reach the primary carnivore level in shortest time.

106. The utilization or consumption efficiency of herbivorous is highest in

(a) plankton communities of ocean waters

(b) mature temperate forests

(d) managed rangelands

Ans. (a)

Sol. Recent experimental studies have suggested that a high diversity of phytoplankton species within a single community results in a greater efficiency of light and nutrient utilisation and a higher standing crop of algal biomass. However, at present, direct experimental evidence that phytoplankton resouce use and carbon fixation is directly linked to the diversity of phytoplankton communities is mainly supported from studies of freshwater and brackish water communities.

107. Which of the following is not an attribute of a species that makes it vulnerable to extinction?

(a) Specialized diet

(b) Low dispersal ability

(d) Variable population density

Ans. (c)

Sol. A number of attributes or traits of life cycle have been proposed as factors determining the speices vulnerability to extinction. Important among these include :

1. Rarity : Abundance of species before fragmentation is an important criteria for determining the vulnerability of the species to extinction. It has been found that after fragmentation, rare understory birds occupy fewer forest fragments per species than common ones. Thus fewer individuals of rare species as compared to the common species are likely to occur in habitat fragments. Such species are prone to extinction sooner than other species.

2. Dispersal Ability : Species capable of migrating between fragments of habitats or between mainland areas and the fragments are able to nullify the effect of small population size. Thus mobile species which are capable of migration with ease and swiftness are not at much risk of being lost.

3. Degree of Specialization : Specialized species usually use and subsist on resources which are distributed in small patches in time and space. These resources are likely to be effected more frequently than others. The reason for rarity of specialized species is the rarity of these resources. Such species are therefore, more prone to extinction. Specialized species may also be vulnerable to successional changes in the habitat fragments and to the breakdown of co-evolved mutualism and food web.

4. Niche Location : Species adopted to or able to tolerate conditions at the interface between different of habitats are less affected by habitat fragmentation than others. These species are actually benefitted by habitat fragmentation. Species intolerant to conditions of life at the itnerface between two types of habitats are more prone to extinction in a relatively shorter time.

5. Population Variability : Species with relatively stable populations are much less vulnerable to extinction than those which undergo large fluctuations in population densities. The population of those species with stable population is unlikely to fall below the critical threshold from which it is difficult to recover.

6. Trophic Status of the Species : Animals at higher trophic levels are usually less abundant and thus more vulnerable to extinction because of rarity.

7. Adult Survival Rate : Species with naturally low adult survivl rates are usually more vunerable to extinction than those species, which have higher adult survival rates.

8. Longivity : Usually long-lived animals are less vulnerable to extinction than those species which have shorter life span.

108. Which of the following is wild relative of wheat?

(a) Triticum monococcum

(b) Triticum compactum

(d) Triticum boeoticum

Ans. (a)

Sol. Bread wheat, Triticum astivum, is a hexaploid comprising an "A" genome from the wild diploid Triticum urartu, a "B" genome most likely from Aegilops speltoides and a "D" genome from Triticum tauschii. The first cultivated wheat was the diploid Triticum monococcum einkorn wheat selected from wild T. monococcum and both wild and cultivated monococcums are reproductively isolated from T. urartu with interspecific hybrids being sterile, although the two wild species have similar morphology.

109. Based on the table given below, which of the following option represents the correct match?

Category Plant species

P. Critically endangered 1. Chromolaena odorata

Q. Vulnerable 2. dipterocarpus grandiflorus

R. Extinct 3. Euphorbia mayurnathanii

S. Invasive 4. Saraca asoka

(a) P-1, Q-4, R-3, S-2

(b) P-2, Q-3, R-4, S-1

(d) P-2, Q-4, R-3, S-1

Ans. (d)

Sol. Dipterocarpus grandiflorus is a critically endangered common medium hardwood tree in South-East Asia and India. Its wood is used to produce good quality charcoal, paper pulp and timber sold under the Keruing designation.

Saraca asoca is a plant belonging to the Caesalpinioideae subfamily of the legume family. As a wild tree, the ashoka is a vulnerable species. It is becoming rarer in its natural habitat, but isolated wild ashoka trees are still to be found in the foothills of the central and eastern Himalayas, in scattered locations of the northern plains of India as well as on the wes coast of the subcontinent near Mumbai.

Euphorbia mayurnathanii is a species of plant in the Euphoribiaceae family. It was endemic to the Palghat Gap in India, but is now believed to tbe extinct in the wild.

Chromolaena odorata is considered an invasive weed of field crops and natural environments in its introduced range. It has been reported to be the most problematic invasive species within protected rainforests in Africa. In Western Africa it prevents regeneration of trees species in areas of shifting cultivation. It affects species diversity in southern Africa. The Plant's flammability affects forests edges. In Sri Lanka it is major weed in disturbed areas and coconut plantations.

110.

With reference to the phylogenetic tree presented above, which of the following statement is true?

(a) Amphibians, reptiles, birds and mammals share a common ancestor.

(b) Birds are more closely related to reptiles than to mammals.

(d) Lampreys and mammals are not related.

Ans. (a)

Sol. Tetrapods are vertebrates that have, or had, four limbs and include all amphibians, reptiles, birds, and mammals.

111. For the following inverterbrate structures/organs, identify their major function and the animal group in which they are found:

Nematocyst (P), Protonephridia (Q), Malpighian Tubules (R), Radula (S)

(a) P – Porifera, Skeletal Support; Q – Mollusca, excretion; R – Insecta, respiration; S – Anthozoa, prey capture

(b) P – Anthozoa, prey capture; Q – Planaria, excretion; R – Mollusca, excretion; S – Insecta, respiration

(d) P – Anthozoa, prey capture; Q – Planaria, excretion; R – Insecta, respiration; S – Mollusca, food processing

Ans. (d)

Sol. Nematocysts deliver the venom through the skin, whereas spirocysts are adhesive and ptychocysts are involved in protection. While Anthozoans have the three types of cnidae, medusozoans (Scyphozoans and Cubozoans) contain only nematocysts. "The planarian excretory system is made up of protonephridia, which are branched organs that are widely distributed throughout the body."Malpighian tubule, in insects, any of the excretory organs that lie in the abdominal body cavity and empty into the junction between midgut and hindgut."The radula, part of the odontophore, may be protruded, and it is used in drilling holes in prey or in rasping food particles from a surface.

112. Following is a cladogram showing phylogenetic relationships among a group of plants:

In the above representation A, B, C and D respectively represent

(a) xylem and phloem, embryo, flower, seed.

(b) embryo, xylem and phloem, seed, flowers.

(d) xylem and phloem, flower, embryo, seed

Ans. (b)

Sol. Embryo is ancestral trait of all these species. Xylem and phloem in ferns, pines and oak. Seeds in pines and oak and flowers in oaks only.

113. Match the following human diseases with their causal organisms

P. Sleeping Sikckness 1. Trypanosoma cruzi

Q. Chagas disease 2. Trypanosoma brucei

R. Elephantiasis 3. Borrelia burgdorferi

S. Lyme disease 4. Wuchereria bancrofti

(a) P-2, Q-4, R-3, S-1

(b) P-1, Q-2, R-4, S-3

(d) P-2, Q-4, R-1, S-3

Ans. (c)

Sol. African Trypanosomiasis, also known as "sleeping sickness", is caused by microscopic parasites of the species Trypanosoma brucei."Chagas disease is caused by the parasite Trypanosoma cruzi."Wuchereria bancrofti is a filarial nematode that is the major cause of lymphatic filariasis (elephantiasis). Lyme disease is caused by the bacterium Borrelia burgdorferi and rarely, Borrelia mayonii.

114. If gypsy moth egg density is 160 at time t and 200 at t + 1, what will be its value at time t + 3, assuming that egg density continues to increase at constant rate?

(a) 250

(b) 280

(d) 390

Ans. (c)

Sol.

Hence, N_t = N₀R^t = 160 × (1.25)³ = 312.5

115. The approximate P : B (Net Production : Biomass) ratios in four different ecosystems (A, B, C, D) are A – 0.29; B – 0.042; C – 16.48; D – 8.2.

The four ecosystems are

(a) A – Ocean; B – Lake; C – Grassland; D – Tropical forest

(b) A – Grassland; B – Tropical forest; C – Ocean; D – Lake

(d) A – Grassland; B – Ocean; C – Lake; D – Tropical forest

Ans. (b)

Sol. NPP is high in coral reef since the high light intensity and warm water allow rapid photosynthesis. Ocean has the maximum net production/biomass . Then comes lake and grasslands and at last tropical forest. In Tropical Rainforests, water, sunlight, and high temperatures are consistent and a dense concentration of plants is present, causing both the GPP and NPP to be very high. So, high levels of productivity increase biomass.

116. The birth rates (b) and death rates (d) of two species 1 and 2 in relation to population density (N) are shown in the graph. Which of the following is not true about the density dependent effects on birth rates and death rates?

(a) Birth rates are density-dependent in species 1 and density independent in species 2.

(b) Death rates are density-dependent in species in both the species.

(d) The density-dependent effects on death rates are similar in both the species.

Ans. (d)

Sol. Density-dependent limiting factors cause a population's per capita growth rate to change—typically, to drop—with increasing population density. One example is competition for limited food among members of a population. Density-independent factors affect per capita growth rate independent of population density. Density-independent factors are not influenced by a species population size.

117. For two species A and B in competition, the carrying capacities and competition coefficients are

K_A = 150 K_B = 200

= 1 = 1.3

According to the Lotka-Voltera model of interspecific competition, the outcome of competition will be:

(a) Species A wins

(b) Species B wins

(d) Both species reach an unstable equilibrium

Ans. (b)

Sol. Unstable equilibria — > 1

These criteria correspond to . The Lotka-Volterra competition model has not only stable equilibria, but also unstable equilibria, when both populations are greater than zero. Although an unstable equilibrium cannot persist, > 1 creates interesting and prbably important dynamics. One of the results is referred to as founder control, where either species can colonize a patch and whichever species gets there first (i.e., the founder) can resist any invader.

118. Which one of the following statements is incorrect?

(a) Loss of genetic variation occurs within a small population is called the genetic load.

(b) The number of deleterious alleles present in the gene pool of a population is called the genetic load.

(d) Inbreeding depression resuts from increased homozygosity for deleterious allees.

Ans. (c)

Sol. The second important effect of genetic erosion is a reduction in levels of heterozygosity. At the population level, reduced heterozygosity will be seen as an increase in the number of individuals homozygous at a given locus. At the individual level, a decrease in the number of heterozygous loci within the genotype of a particular plant or animal will occur. As we have seen, loss of heterozygosity is a common consequence of reduced population size. Alleles may be lost through genetic drift or because individuals carrying them die without reproducing. Smaller populations also increase the likelihood of inbreeding, which inevitably increase homozygosity.

119. During which of the following major mass extinction events, over 95% of the marine species disappeared from the planet Earth?

(a) Ordovician

(b) Devonian

(d) Triassic

Ans. (c)

Sol. The Permian Period was the final period of the Paleozoic Era. Lasting from 299 million to 251 million years ago, it followed the Carboniferous Period and preceded the Triassic Period.

The Permian extinction was characterized by the elimination of over 95 percent of marine and 70 percent of terrestrial species. In addition, over half of all taxonomic families present at the time disappeared. This event ranks first in severity of the five major extinction episodes that span geologic time.

120. Which of the following global hotspots of biodiversity has the highest number of endemic palnts and vertebrates?

(a) Sundaland

(b) Tropical Andes

(d) Mesoamerican forests

Ans. (b)

Sol. The montane and pre-montane forests of northern Peru lie at the heart of the Tropical Andes Biodiversity Hotspot and are among the most threatened forested areas in the world.

The tropical Andes are home to incredible levels of biodiversity with ~30,000 vascular plant species, 50% of which are endemic and the highest number of vertebrate species of any "Biodiversity Hotspot". This includes 584 species and 69 genera of endemic birds.

Diversity and endemism of mammals is similarity high with at least 75 species and five monotypic genera endemic to the area.

121. Fossils of the same species of freshwater reptiles have found in South America and Africa. Based on the current understanding, which of the following is the best possible explanation for this pattern?

(a) The same species which of the following is the best possibe explanation for this pattern?

(b) Species migrated from Africa to establish new populations in South America.

(d) South America and Africa were joined at some point in Earth's history.

Ans. (d)

Sol. Trained as a meterologist, Wegner was intrigued by the interlocking fit of Africa's and South America's shorelines. Wegener then assembled an impressive amount of evidence to show that Earth's continents were once connected in a single supercontinent.

Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces. For example, the Appalachian Mountains (United States) and Caledonian Mountains (Scotland) fit together, as do the Karroo strata in South Africa and Santa Catarina rocks in Brazil.

122. Match major events in the history of life Earth's geological period.

Event Geological period

P. First reptiles 1. Quaternary

Q. First mammals 2. Tertiary

R. First humans 3. Cretaceous

S. First amphibians 4. Triassic

5. Carboniferous

6. Devonian

(a) P-5, Q-1, R-2, S-5

(b) P-5, Q-4, R-1, S-6

(d) P-3, Q-1, R-6, S-5

Ans. (b)

Sol.

123. Fruit colour of wild Solanum nigrum is controlled by two alleles of a gene (A and a). The frequency of A, p = 0.8 and a, q = 0.2. In a neighbouring field a tetraploid genotype of S. nigrum was found. After critical examination five distinct genotypes found; which aer AAAA, AAAa, AAaa, Aaaa and aaaa. Following Hardy Weinberg principle and assuming the same allelic frequency as that of diploid population, the numbers of phenotypes calculated within a population of 1000 plansts arer close to one of the following:

AAAA : AAAa : AAaa : Aaaa : aaaa

(a) 409 : 409 : 154 : 26 : 2

(b) 420 : 420 : 140 : 18 : 2

(d) 409 : 420 : 144 : 25 : 2

Ans. (a)

Sol. From Hardy-Weinberg principle for tetraploid genotype : (p + q)⁴ = p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴

Number of phenotypes of AAAA = p⁴ × 1000 = (0.8)⁴ × 1000 = 0.4096 × 1000 = 409

Number of phenotypes of AAAa = 4p³q × 1000 = 4 × (0.8)⁴ × (0.2) × 1000 = 409

Number of phenotypes of AAaa = 6p²q² × 1000 = 6 × (0.8)² × (0.2)² × 1000 154

Number of phenotypes of Aaaa = 4pq³ × 1000 = 4 × 0.8 × (0.2)³ × 1000 26

Number of phenotypes of aaaa = q⁴ × 1000 = (0.2)⁴ × 1000 2

124. In a lake subjected to progressive eutrophication, temporal changes in the magnitude of selected parameters (P, Q, R, s) are shown in the graph

The parameters P, Q, R, S are

(a) P – Green algal biomass, Q – Cyanobacterial biomass, R – Dissoved Oxygen concentration, S – Biological Oxygen Demand

(b) P – Biological Oxygen Demand, Q – Cyanobacterial biomass, R – Dissoved Oxygen concentration, S – Green algal biomass

(d) P – Cyanobacterial biomass, Q – Biological Oxygen Demand, R – Green algal biomass, S – Dissoved Oxygen concentration

Ans. (b)

Sol. Biochemical oxygen demand is the amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period. "Eutrophication sets off a chain reaction in the ecosystem, starting with an overabundance of algae and plants. The excess algae and plant matter eventually decompose, producing large amounts of carbon dioxide. This lowers the pH of seawater, a process known as ocean acidification. "One of the negative impacts of eutrophication and increased algal growth is a loss of available oxygen, known as anoxia. These anoxic conditions can kill fish and other aquatic organisms such as amphibians. So, dissolved oxygen concentration and green algal biomass will decrease with progressive eutrophication and biological oxygen demand and cynobacterial biomass will increase.

125. Consider an autosomal locus with two alleles A₁ and A₂ at frequencies of 0.6 and 0.4 respectively. Each generation, A₁ mutates to A₂ at a rate of µ = 1 × 10^–5 while A₂ mutates to A₁ at a rate of v = 2 × 10^–5. Assume that the population is infinitely large no other evolutionary force is acting. The equilibrium frequency of allele A₁ is

(a) 1.0

(b) 0.5

(d) 0.33

Ans. (c)

Sol. The equilibrium frequency of allele A₁ =

126. With reference to the graph given below, identity the optimal territory size.

(a) P

(b) Q

(d) S

Ans. (a)

Sol. P is the optimum territory size due to presence of less cost and high benefit.

127. A particular behavioural variant affects fitness of an organism. The relationship between the ferquency of the variant in the population and fitness are plotted below. In which of these cases is the behavioural variant most likely to reach a frequency of 1?

(a) only Q

(b) onlyR

(d) P and S

Ans. (b)

Sol. In graph R, behavioral variant is most likely to reach a frequency of 1, where the fitness of behaviour directly increase with frequency of variant.

128. The coefficient of relatedness between individuals A and B, A and D, and between D and C is

(a) 0.5, 0.25, 0.125 respectively

(b) 0.5, 0.5, 0.25 respectively

(d) 0.125, 0.5, 0.5 respectively

Ans. (a)

Sol. Coefficient of relatedness between common ancestor = r = ; where, n = number of step in genealogy

129. One hundred independent populations of Drosophila are established with10 individuals in each population, of which, one individual isof Aa genotype and the other nine are of AA genotype. If random genetic drift is the only mechanism acting on these populations, then, after a large number of generations, the expected number of populations fixed for the 'a'allele is

(a) 75

(b) 50

(d) 5

Ans. (d)

Sol. Genetic drift is a mechanism of evolution in which allele frequencies of a population change over generations due to chance. After large number of generations expected number of population in which allele would have been fixed = no. of population/ 2N

So, 100/2 × 10 = 5

130. TILLING is a reverse genetics approach used in functional genomics.Which one of the following is used for TILLING?

(a) T-DNA tagging by Agrobacterium-mediated transformation.

(b) Transposon tagging using Ac/Ds elements.

(d) Protoplast transformation by electroporation.

Ans. (c)

Sol. Tilling populations are generated by mutagenesis. Mutagens are forms of energy or chemical substances that increase the frequency of mutations above that of background in the genomes of exposed organisms. Since the 1920s, several mutagens have been used to induce genetic variation and to develop mutant populations. For tilling, ehtyl methane sulphonate (EMS) has become the most commonly used because of its ease of handling, well established mode of action and its effectiveness in inducing a high frequency of point mutations in the absence of gross chromosomal abnormalities.

131. Which one of the following will be observed when auxin to cytokinin ratio is increased in the culture medium during organogenesis from tobacco pith callus?

(a) Adventitious roots will form

(b) Adventitious shoot will form

(d) There will be no shoot formation

Ans. (a)

Sol. Among phytohormones, auxin plays an essential role in regulating roots development and it has been shown to be intimately involved in the process of adventitious rooting.

132. A and B are two enantiomeric helical peptides. Their charity can be determined by recording their

(a) circulardichroism spectrum

(b) UV spectrum

(d) Edman spectrum

Ans. (a)

Sol. Edman degradation, developed by Pehr Edman, is a method of sequencing amino acids in a peptide. In this method, the amino terminal residue is labelled and cleaved from the peptide without disrupting the peptide bonds between other amino acid residues.

Because the Eman degradation proceeds from the N-terminus of the protein, it will not work if the N-terminal amino acid has been chemically modified or if it is concealed within the body of the protein. It also requires the use of either guesswork or a separate preocedure to determien the position of disulphide bridges and peptide concentrations of 1 picomolar or above for discermible results.

133. The use of Krusal Wallis test is most appropriate in which of these cases?

(a) There are more than two groups and each group is normally distributed.

(b) There are more than two groups and the distribution in each group is not normal.

(d) There are two groups and the distribution in each group is not normal.

Ans. (b)

Sol. The Kruskal-Wallis test is similar to an ANOVA test and is used to compare more than two groups of independent, ordinal or non-normally distributed data. If acetaminophen is added as a third group to the fever-reducing study and self-assessment results are compared between the groups, a Kruskal-Wallis test could be used to evaluate the results. Similar to ANOVA, a post hoc analysis is necessary to determien where the difference lies.

134. Which one of the following can be analysed using Surface Plasmon Resonance method?

(a) Radiolabelled DNA probes

(b) Protein structure

(d) Label-free bimolecular interaction

Ans. (d)

Sol. Surface Plasmon Resonance (SPR) is a label-free detection method by which molecular interactions may be analyzed on a surface. Binding data are collected in real time, allowing the determination of interaction kinetics. SPR imaging (SPRi) the focus of this review, improves upon the efficiency of SPR by facilitating analysis of multiple interactions simultaneously.

135. Which one of the following statements is correct for amplified-fragment length poymorphism (AFLP)?

(a) PCR using a combinaiton of random and gene-specific primers.

(b) PCR amplification followed by digestion with restriction enzymes.

(d) Digestion of DNA with restriction enzymes followed by two PCR step.

Ans. (d)

Sol. Amplified Fragment Length Polymorphism (AFLP) is a polymerase chain reaction (PCR) based genetic fingerprinting technique. AFLP usese restriction enzymes to cut genomic DNA, followed by ligation of comlementary double stranded adaptors to the ends of the restriction fragments. A subset of the restriction fragments are then amplified using two primers complementary to the adaptor and restriction site fragments. The fragments are visualized on denautring polyacrylamide gels either through autoradiographic or fluorescence methodologies. AFLP-PCR is a highly sensitive method for detecting polymorphisms in DNA. The procedure of this technique is divided into three steps :

Digestion of total cellular DNA with one or more restriction enzymes and ligation of restriction enzymes and ligation of restriction half-site specific adaptors to all restriction fragments.

Selective amplification of some of these fragments with two PCR primers that have corresponding adaptor and restriction site-specific sequences.

Electrophoretic separation of amplicons on a gel matrix, followed by visualization of the band pattern.

136. In an effort to produce gene knockout mice, a gene targeted homologous recombination was tried with the exogenous DNA containing neo^rgene (confer G-418 resistance) and tk^Hsvgene (confers sensitivity to the cytotoxic nucleotide analog ganciclovir). If the neo^r gene was inserted within the target gene in theexogenous DNA and considering that both homologous and non-homologous recombination (random integration) is taking pace, which one of the following statements is not correct about the possibel outcome of the experiment?

(a) Cells with non-homologous insertion will be sensitive to ganciclovir.

(b) Non-recombinant cells will be sensitive towards G-418 and resistant to ganciclovir.

(d) Homologous recombination will grow in G-418 containing media but will be sensitive towards ganciclovir.

Ans. (d)

Sol. The targeting vector is designed to contain both neomycin-resistant and ganciclovir-sensitive (TK) sequences. Currently, the neomycin-resistance gene, called NeoR, is a popular marker gene of choice for generating knockout mice. If random insertion occurs, both the NeoR gene and the TK gene are inserted into the genome. As a result, the cells are resistant to neomycin, but they die in the presence of ganciclovir. In comparison, when the targeted gene segment is correctly replaced, the TK gene is not inserted into the chromosome along with the NeoR gene, so the resultant cells are resistant to both neomycin and ganciclovir. Therefore, the presence of the TK gene in the targeting vector allows researchers to efficiently screen for mouse cells that have correctly replaced the targeted gene segment by growing these cells in the presence of both neomycin and ganciclovir.

137. In a typical gene cloning experiment, by mistake a researcher introduced the DNA of interest within ampicillin resistant gene instead of lac Z gene. The competent cells were allowed to take up to the plasmid and then plated in the media containing ampicillin, X-gal and IPTG and subjected to blue-white screening. Considering all plasmids were recombinant which one of the following statements correctly describes theoutcome of the experiment?

(a) The bacteria which took up the plasmids would grow and give blue colonies.

(b) The bacteria which took up the plasmids would not grow.

(d) All of the bacteria would grow and give white colonies.

Ans. (b)

Sol. As ampicillin resistant gene is inactivated by DNA of interest. So the cells which took up the plasmid will not grow in presence of ampicillin.

138. The following table shows the mean and variance of population densities of species A, B and C.

Statistic Species A Species B Species C

5.3 7.05 5.30

Variance s² 5.05 0.35 50.5

Based on the above, which of the following statements is correct?

(a) Species A and B show uniform distribution, whereas species C shows clumped distributed.

(b) Species A shows random distribution, species B shows uniform distribution, and species C shows clumped distribution.

(d) Species A shows clumped distribution, species B shows random distribution, and species C shows uniform distribution.

Ans. (b)

Sol. The species will show the uniform distribution, if the ratio of variance and maen tends to zero (0).

Uniform distribution = = 0.049

The species will show the random distribution, if the ratio of variacne and mean tends to one (1).

Random distribution = = 0.95

The species will show the Clustered (Clumped) distribution, if the ratio of variance and mean tends to infinity,

Clustered (Clumped) distribution = = 9.52

139. Performance of biosensor is evaluated by their response to the presence of an analyte. The physiological relevant concentration of analyte is between 10 µM and 50 µM. Which among the following biosensor respnses is best?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. A substance of interest that needs detection. For instance, glucose is an 'analyte' in a biosensor designed to detect glucose. Bioreceptor: A molecule that specifically recognises the analyte is known as a bioreceptor. The limit of detection is expressed in units of concentration and, following IUPAC definition, indicates the smallest solute concentration than a given analytical system is able to distinguish with a reasonable reliability of a sample without analyte. C biosensor response is best for concentration of analyte between 10microM to 50microM.

140. Molecular polymorphic markers are already known with respect to tobacco mosaic virus (TMV) resistance in tobacco. Among these, which marker system you will select that will be simple, economic and less time consuming?

(a) RAPD

(b) RFLP

(d) EST-SSR

Ans. (d)

Sol. Expressed sequence tag-derived simple sequence repeat markers (EST-SSRs) are the markers of choice, because they are abundant, co-dominant, highly polymorphic, and are easily transferable among phylogenetically related species.

141. The sequence of the peptide KGLITRTGLIKR can be unequivocally determined by

(a) Only Edman degradation

(b) Amino acid analysis and MALDI MS/MS mass spectrometry.

(d) MALDI mass spectrometry after treatment of the peptide with trypsin.

Ans. (a)

Sol. The emphasis on the word 'unequivocally' mena that "something that doesn't leave a doubt". Although it is possible to determine the sequence of a peptide by doing amino acid analysis, MALDI MS/MS. However, these techniques provide the sequence with some confidence score sya 78% or any other value (i.e. it cannot be determined with 100% surely that sequence is correct). But using a true protein sequencing technique called Edman degradation method can provide the actual protein sequence without leaving a doubt. (i.e. with more confidence). As we know that MALDI alone without MS/MS it is not possible to determine protein sequence.

142. From statements on protein structure and interactions detailed below, indicate the correct statement:

(a) The concentration of a tryptophan containing protein can be determined by monitoring the fluorescence spectrum of the protein.

(b) A peptide with equal number of Glu and Lys amino acids can show multiple charged species in its electrospray ionization mass spectrum.

(c) The circular dichroism spectrum of a protein shows predominantly helical conformation. Analysis of its two dimensional NMR spectrum shows predominantly -structure.

(d) Binding constant can be determined by two interacting moecules by the technique of surface plasmon resonance only if there is strong hydrophobic interactions between them.

Ans. (b)

Sol. ESI-MS is the process through which proteins, or macromolecules, in the liquid phase are charged and fragmented into smaller aerosol droplets. Peptide with equal number of Glu and Lys amino acids will show multiple charged species in its ESI- mass spectrum.

143. Radioimmunoassay (RIA) can be employed for the detection of insulin in blood plasma. For this, ¹²⁵I-labelled insulin is mixed and allowed to bind with a known concentration of anti-insulin antibody. A known volume of patients' blood plasma is then added to the conjugate and allowed to compete with the antigen binding sites of antibody. The bound antigen is then separated from unbound ones and the radioactivity of free antigen is then measured by gamma counter. Following are some of the statements made about this assay.

P. The ratio of radioactive count for unbound antigen to the bound one is more at the end of reaction.

Q. The ratio of radioactive count for unbound antigen to the bound one is less at the end of reaction.

R. For a diabetic patient, the radioactive count for free antigen is less than that for a normal individual.

S. For a diabetic patient, the radioactive count for free antigen is more than that for a normal individual.

Which of the above statements are true?

(a) P and R

(b) P and S

(d) Q and S

Ans. (a)

Sol. A radioimmunoassay (RIA) is an immunoassay that uses radiolabeled molecules in a stepwise formation of immune complexes. A RIA is a very sensitive in vitro assay technique used to measure concentrations of substances, usually measuring antigen concentrations (for example, hormone levels in blood) by use of antibodies. Insulin radioimmunoassay (RIA) is a double-antibody batch method. Insulin in the specimen competes with a fixed amount of 125I-labelled insulin for the binding sites of the specific insulin antibodies. Bound and free insulin are separated by adding a second antibody, centrifuging, and decanting.

144. It is hypothesized that the mean (µ₀) dry weight of a female in a Drosophia population is 4.5 mg. In a sample of 16 female with = 4.8 mg and s = 0.8 mg, what dry weight values would lead to rejection of the null hypothesis at p = 0.05 level? (take t_0.05 = 2.1)

(a) Values lower than 4.0 and values higher than 5.6

(b) Values lower than 3.20 and values higher than 6.40

(d) Values lower than 3.22 and values higher than 6.48

Ans. (c)

Sol. A confidence interval is a specific interval estimate of a parameter determined by using data obtained from a sample any by using the specific confidence level of the estimate. It is hypothesized that the mean (µ₀) dry weight of a female in a Drosophila population is H₀ : µ = 4.5 mg. The central limit theorem states that when the sample size is large, approximately 95% of the sample maens taken from a population and same sameple size will fall within + 2.1 standard errors of the population mean, that is, .

Now, if a specific sample mean is selected, say , there is a 95% probability that the interval contains . Likewise, there is a 95% probability that the interval specified by .

For rejection of the hypothesis, = sample mean

= sample mean, s = standard deviation and n = size of the sample.

The value used for the 95% confidence interval of t* is 2.1

Where, L = lower estimate of the population and U = upper estimate of the population.

= 4.38, 5.22

Hence, one can say with 95% confidence that the interval between 4.38 and 5.22 dry weight of a female in a Drosophila population does not contain the population mean, based on a sample of 16 female.

145. A researcher wants to obtain complete chemical information, i.e. head groups and fatty acids of phospholipids from liver tissues. Phospholipids have fatty acids of different lengths and unsaturation and also the head groups are of different chemistries. Which of the following combination of techniques would provide complete chemical description of phospholipids?

(a) Only thin layer chromatography (TLC)

(b) TLC and gas chromatography

(d) Only paper chromatography

Ans. (b)

Sol. TLC" stands for "thin layer chromatography" while "GLC" is short for gas-liquid chromatography. Both techniques are capable of separating the components of a mixture, identifying a compound, determining the purity of a substance, and monitoring the progress of reaction of the mixture. Gas chromatography is usually used to separate and measure organic molecules and gases. For the technique to function, the components being analyzed must be volatile, be thermally stable, and have a molecular weight of below 1250 Da.

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CSIR NET BIOLOGY (JUNE - 2016)
Previous Year Question Paper with Solution.