Gate life sciences 2009


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This ppt contains few solved questions of GATE 2009 examination along with explanations. This will be helpful for all those who are preparing for GATE, CSIR, UGC NET, etc. Complete set of questions along with answers and explanations can be viewed at

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Gate life sciences 2009

  1. 1. GATE LIFE SCIENCES 2009 Solved by Anna Purna Answers reinforced with explanations http://
  2. 2. Q.1 Which of these can be an antigen but cannot induce immune responses? (A) Hemocyanin (B) Influenza virus (C) Azobenzene arsonate (D) Corynebacteria Ans C P-Azobenzenearsonate is a hapten which capable of eliciting both antibody formation and delayed hypersensitivity when bound to aromatic amino acids, polypeptides or proteins. Haptens can elicit an immune response only when attached to a large carrier.
  3. 3. Q.2 Generally, the rate-limiting step of major metabolic pathways is a reaction (A) in which the availability of the substrate is limited. (B) catalyzed by an allosteric enzyme. (C) catalyzed by an enzyme with very low Km. (D) whose products are not readily consumed by the subsequent step of the pathway. Ans C The rate-limiting step of every metabolic pathway is far from equilibrium because of the relatively low activity of the enzyme that catalyzes it. The rate of this reaction is not limited by substrate availability, but only by the activity of this enzyme. Low Km of enzyme indicates a high affinity between enzyme and substrate, meaning that the rate will approach V max more quickly.
  4. 4. Q.3 Equal volumes of two buffers of pH 4 and pH 6 of identical ionic strengths are mixed. The resultant pH is (A) close to 4. (B) close to 5. (C) close to 6. (D) exactly 5. Ans A The hydrogen ion concentration ( [H+] concentration (mol/l) ) in a solution of pH 4 is 0.0001 (1 x 10 -4 ). Similarly the hydrogen ion concentration in a solution of pH 6 is 0.000001( 1x10 -6 ). When you add two solutions of identical ionic strengths (you get a mixture of all ions in the solution), the sum is equal to (0.0001 + 0.000001 = 0.000101). 0.000101 is closer to 0.0001 and hence the pH is closer to 4.
  5. 5. Q.4 The formation of ATP from ADP and Pi is not a spontaneous reaction. A reason for this is (A) ATP readily ionizes. (B) electrostatic repulsion in ATP is lower than that in ADP. (C) ATP is better hydrated than the total hydration levels of ADP and Pi. (D) resonance stabilization of P-O bonds in Pi is higher than that in ATP. ATP is an unstable molecule in and gets hydrolysed to ADP and phosphate easily. This is because the strength of the bonds between the phosphate groups in ATP are less than the strength of the hydrogen bonds (hydration bonds), between its products (ADP + phosphate), and water. Ans D
  6. 6. Q.5 A beam of light passes through 1 cm of a colored solution. Eighty percent of the incident light is transmitted. If the incident light passes through 2 cm of the same solution, the percentage of transmitted light is (A) 60 (B) 64 (C) 70 (D) 40 Ans A According to the Law of absorption the amount of light absorbed is proportional to the thickness of the absorbing material and is independent of the intensity of the incident light.
  7. 7. Q.6 Kyoto Protocol is related to (A) Acid rain (B) Photochemical smog (C) Ozone hole (D) Global warming Ans D Kyoto Protocol sets binding targets for 37 industrialized countries and the European community for reducing greenhouse gas (GHG) emissions which in turn lead to global warming.
  8. 8. Q.7 During receptor-mediated endocytosis of LDL bound to its receptor (A) both receptor and ligand are degraded, (B) the receptor is degraded and the ligand is recycled. (C) both are recycled. (D) the ligand is degraded and the receptor is recycled. Ans D The receptor is recycled and returned to the plasma membrane to collect more LDL particles where as the ligand or LDL is broken down to cholesterol, amino acids and fatty acids.
  9. 9. Q.8 Which of the following statements are INCORRECT ? P. The frequency of recombination is a measure of linkage between genes on the same chromosome. Q. DNA polymerase I is the true DNA replicase in E.coli. R. The conserved element closest to the transcription initiation site is called the CAAT box. S. The introns in the nuclear pre-mRNAs are excised by spliceosomes. (A) P. Q (B) Q. R (C) P. R (D) P. S Ans B
  10. 10. Q.9 <ul><li>(A) P-1, Q-5, R-6, S-3 </li></ul><ul><li>(B) P-2, Q-6, R-3, S-4 </li></ul><ul><li>(C) P-1, Q-6, R-4, S-2 </li></ul><ul><li>(D) P-1, Q-2, R-6, S-4 </li></ul>Ans B
  11. 11. Q.10 Density of cells of a bacterial culture is routinely measured using Spectrophotometer. This is based on the principle of (A) Light absorption (B) Light diffraction (C) Light scattering (D) Light reflection Ans A A spectrophotometer is used to measure the amount of light a sample (be it liquid or solid) absorbs.
  12. 12. Q.11 Pseudopeptidoglycan is present in the cell wall of (A) Escherichia coli (B) Bacillus subtilis (C) Saccharomyces cerevisiae (D) Methanococcus jannaschii Ans D Pseudopeptidoglycan is a major cell wall component of some archaea. The basic components of Pseudopeptidoglycan are N-acetylglucosamine and N-acetyltalosaminuronic acid (Peptidoglycan has N -acetylmuramic acid instead), which are linked by a β-1,3-glycosidic bond.
  13. 13. Q.12 A silent mutation is one that (A) results in a truncated polypeptide (B) replaces an amino acid with an equivalent amino acid in a polypeptide (C) does not change the amino acid sequence of the polypeptide (D) changes the reading frame of the mRNA leading to an altered polypeptide Ans C Silent mutations are DNA mutations that do not result in a change to the amino acid sequence of a protein. They may occur in a non-coding region (outside of a gene or within an intron), or they may occur within an exon in a manner that does not alter the final amino acid sequence.
  14. 14. Q.13 RecA is a protein involved in (A) Recombinational repair (B) Mismatch repair (C) Nucleotide excision repair (D) Base excision repair Ans A RecA is a protein in E. coli involved in recombinational repair of damaged DNA and in SOS repair. RecA catalyzes strand pairing, or strand assimilation-the joining of two different DNAs by homologous base pairing with each other.
  15. 15. Q.14 Effective chemotherapeutic agents are difficult to develop for the treatment of fungal infections because (A) Fungi have cell wall. (B) Fungi have better mechanisms to inactivate drugs. (C) Fungi are eukaryotic cells and their cellular machinery is similar to that of the host. (D) Fungal pathogens typically infect organs inaccessible for antibiotic treatment. Ans A Fungi are eukaryotic and are composed of rigid cell wall largely made of chitin which is a polymer of N-acetylglucosamine rather than peptidoglycan which is a characteristic component of most bacterial cell walls.
  16. 16. Q.15 During hibernation in a hibernating mammal, its body temperature would be (A) Lower than normal state (B) Same as normal state (C) Higher than normal state (D) Fluctuate between high and low points Ans A Hibernation is a state of inactivity and metabolic depression in animals, characterized by lower body temperature, slower breathing, and lower metabolic rate.
  17. 17. Q.16 Mendel‘s principle of segregation means that the germ cells (egg or sperm) always receive (A) One of the paired alleles (B) One pair of alleles (C) One quarter of the genes (D) Any pair of alleles Ans A The Law of Segregation states that every individual possesses a pair of genes for any particular trait and that each parent passes a randomly selected copy of only one of these to its offspring through eggs or sperms.
  18. 18. Q.17 Anabolic steroids, taken illegally by sportspersons to enhance their physical strength, are synthetic analogues for natural _______________ from the ___________. (A) testosterone .....,. anterior pituitary (B) FSH and LH ....... posterior pituitary (C] cortisol ..,....... thyroid (D) androgen ......... gonads Ans A They are drugs which mimic the effects of the male sex hormones: testosterone and dihydrotestosterone. They increase protein synthesis within cells, which results in the buildup of cellular tissue (anabolism), especially in muscles.
  19. 19. Q.18 Homology in anatomical parts helps in determining evolutionary kinship because (A) Homologous body parts invariably perform similar functions (B] Display evolutionary adaptations (C] Undergo similar genetic changes (D) Have common embryological origin Ans D Homologous structures are similar in different species because the species have common descent. They may or may not perform the same function. An example is the forelimb structure shared by cats and whales.
  20. 20. Q.19 The human immune system is able to mount a response when it encounters a novel microorganism for the first time because (A) White blood cells are able to change their antigen specificity depending upon the microorganism they interact with. (B) Our body contains millions of different kinds of white blood cells, each with a unique type of antigen receptor. (C) Bone marrow cells make different antigen receptors depending upon the kind of invading microorgamsrn. (D) Bone marrow cells are able to change their antigen specificity upon physical interaction with the microorganism. Ans C The clonal selection hypothesis states that the germline encodes many different antigen receptors - one for each antigenic determinant to which an individual will be capable of mounting an immune response.
  21. 21. Q.20 To produce plants that are homozygous for all traits, the best choice is (A) Protoplatst culture (B) Cell suspension culture (C) Anther and pollen culture (D) Apical meristem culture Ans C Anther and pollen culture helps to produce haploid plants containing a single set of chromosomes. Haploid plants can be used to produce homozygous diploid or polyploid plants which are valuable for breeding.
  22. 22. Thank You