CSIR NET Life Science Coaching

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30/03/2025

"๐‚๐’๐ˆ๐‘ ๐๐„๐“ ๐‹๐ข๐Ÿ๐ž ๐’๐œ๐ข๐ž๐ง๐œ๐ž ๐๐ฎ๐ข๐ณ ๐‚๐ก๐š๐ฅ๐ฅ๐ž๐ง๐ ๐ž! ๐Ÿงฌ
Test your knowledge with these 25 critical thinking MCQs from Unit 1 of CSIR NET Life Science! ๐Ÿ’ก๐ŸŽฏ Drop your answers in the comments and challenge your friends. Letโ€™s see who gets the highest score! ๐Ÿ“š

Stay tuned for the answers and explanations!

CSIR NET Life Science โ€“ Unit 1: MCQ Quiz

Total Questions: 25 | Marks: 50 (2 marks each)

1. Which of the following interactions contributes most significantly to the stability of DNA double helix?
a) Hydrogen bonding
b) Ionic bonding
c) Hydrophobic interactions
d) Covalent bonding

2. What happens to the pKa of a weak acid when it is placed in a highly polar solvent?
a) Increases
b) Decreases
c) Remains unchanged
d) Becomes zero

3. Which of the following is NOT a non-covalent interaction in biological systems?
a) Peptide bond formation
b) Van der Waals forces
c) Hydrophobic interactions
d) Hydrogen bonding

4. Which property of water makes it a suitable medium for biochemical reactions?
a) Low dielectric constant
b) High heat capacity
c) High viscosity
d) Low surface tension

5. The strongest chemical bond found in biological macromolecules is:
a) Hydrogen bond
b) Ionic bond
c) Covalent bond
d) Van der Waals force

6. Which type of amino acid side chain is most likely to be found in the interior of a globular protein?
a) Hydrophilic
b) Hydrophobic
c) Positively charged
d) Negatively charged

7. Which of the following sugars is a reducing sugar?
a) Sucrose
b) Cellulose
c) Maltose
d) Raffinose

8. In a ฮฒ-sheet, the hydrogen bonds are:
a) Parallel to the polypeptide backbone
b) Perpendicular to the polypeptide backbone
c) Absent
d) Responsible for disulfide bridge formation

9. Phospholipids form bilayers in aqueous solutions due to their:
a) Hydrophilic tails
b) Amphipathic nature
c) Hydrogen bonding between phosphate groups
d) Covalent interactions

10. Why is ATP considered an energy-rich molecule?
a) High-energy phosphate bonds
b) High content of nitrogen
c) Its ability to form hydrogen bonds
d) It is a nucleoside derivative

11. Which of the following is NOT a property of enzymes?
a) They increase activation energy
b) They remain unchanged after the reaction
c) They are highly specific
d) They lower activation energy

12. The substrate concentration at which an enzyme operates at half of its maximum velocity is known as:
a) Vmax
b) Km
c) Kcat
d) Ki

13. Competitive inhibition of an enzyme leads to:
a) Decreased Vmax and unchanged Km
b) Increased Vmax and increased Km
c) Unchanged Vmax and increased Km
d) Increased Vmax and decreased Km

14. Allosteric enzymes:
a) Follow Michaelis-Menten kinetics
b) Show cooperative binding
c) Are not affected by inhibitors
d) Work only at high temperatures

15. If an enzyme is saturated with substrate, how can its activity be further increased?
a) Adding more substrate
b) Increasing the temperature indefinitely
c) Increasing enzyme concentration
d) Increasing product concentration

16. Which of the following correctly describes the Gibbs free energy (ฮ”G) for a spontaneous reaction?
a) ฮ”G > 0
b) ฮ”G < 0
c) ฮ”G = 0
d) ฮ”G is positive and entropy increases

17. Which metabolic pathway directly generates ATP without involvement of the electron transport chain?
a) Glycolysis
b) Oxidative phosphorylation
c) Fatty acid oxidation
d) Citric acid cycle

18. Uncoupling of oxidative phosphorylation leads to:
a) Increased ATP production
b) Heat generation without ATP synthesis
c) Complete inhibition of the electron transport chain
d) Decreased oxygen consumption

19. A mutation in the ATP synthase enzyme would likely affect:
a) Glycolysis
b) ATP production in mitochondria
c) Krebs cycle
d) Allosteric regulation of glycolysis

20. The total ATP yield from complete oxidation of one molecule of glucose is approximately:
a) 2 ATP
b) 12 ATP
c) 32 ATP
d) 100 ATP

21. SDS-PAGE separates proteins based on:
a) Charge
b) Size
c) Hydrophobicity
d) Affinity for SDS

22. Which of the following is the first step in western blotting?
a) Protein separation by SDS-PAGE
b) Antibody incubation
c) Substrate detection
d) Membrane blocking

23. Which of the following chromatography techniques is best suited for purifying a protein based on charge?
a) Gel filtration
b) Ion-exchange
c) Affinity chromatography
d) Thin-layer chromatography

24. The primary advantage of NMR spectroscopy over X-ray crystallography for protein structure determination is:
a) Ability to study proteins in solution
b) Higher resolution of structures
c) Requirement of large crystals
d) More straightforward data analysis

25. Which technique is most suitable for determining the molecular weight of a protein?
a) UV-Vis spectroscopy
b) Mass spectrometry
c) X-ray crystallography
d) IR Spectroscopy

30/03/2025

๐™๐™๐™ž๐™จ ๐™„๐™ฃ๐™จ๐™ฉ๐™ž๐™ฉ๐™ช๐™ฉ๐™š ๐™ž๐™จ ๐™– ๐™™๐™š๐™™๐™ž๐™˜๐™–๐™ฉ๐™š๐™™ ๐™˜๐™ค๐™–๐™˜๐™๐™ž๐™ฃ๐™œ ๐™˜๐™š๐™ฃ๐™ฉ๐™š๐™ง ๐™›๐™ค๐™ง ๐˜พ๐™Ž๐™„๐™ ๐™‰๐™€๐™ ๐™‡๐™ž๐™›๐™š ๐™Ž๐™˜๐™ž๐™š๐™ฃ๐™˜๐™š ๐™–๐™จ๐™ฅ๐™ž๐™ง๐™–๐™ฃ๐™ฉ๐™จ, ๐™ค๐™›๐™›๐™š๐™ง๐™ž๐™ฃ๐™œ ๐™š๐™ญ๐™ฅ๐™š๐™ง๐™ฉ ๐™œ๐™ช๐™ž๐™™๐™–๐™ฃ๐™˜๐™š, ๐™˜๐™ค๐™ข๐™ฅ๐™ง๐™š๐™๐™š๐™ฃ๐™จ๐™ž๐™ซ๐™š ๐™จ๐™ฉ๐™ช๐™™๐™ฎ ๐™ข๐™–๐™ฉ๐™š๐™ง๐™ž๐™–๐™ก๐™จ, ๐™–๐™ฃ๐™™ ๐™ง๐™š๐™จ๐™ช๐™ก๐™ฉ-๐™ค๐™ง๐™ž๐™š๐™ฃ๐™ฉ๐™š๐™™ ๐™จ๐™ฉ๐™ง๐™–๐™ฉ๐™š๐™œ๐™ž๐™š๐™จ. ๐™Š๐™ช๐™ง ๐™œ๐™ค๐™–๐™ก ๐™ž๐™จ ๐™ฉ๐™ค ๐™๐™š๐™ก๐™ฅ ๐™จ๐™ฉ๐™ช๐™™๐™š๐™ฃ๐™ฉ๐™จ ๐™š๐™ญ๐™˜๐™š๐™ก ๐™ž๐™ฃ ๐™ฉ๐™๐™š ๐™š๐™ญ๐™–๐™ข ๐™ฉ๐™๐™ง๐™ค๐™ช๐™œ๐™ ๐™จ๐™ฉ๐™ง๐™ช๐™˜๐™ฉ๐™ช๐™ง๐™š๐™™ ๐™˜๐™ค๐™ช๐™ง๐™จ๐™š๐™จ, ๐™ข๐™ค๐™˜๐™  ๐™ฉ๐™š๐™จ๐™ฉ๐™จ, ๐™–๐™ฃ๐™™ ๐™ฅ๐™š๐™ง๐™จ๐™ค๐™ฃ๐™–๐™ก๐™ž๐™ฏ๐™š๐™™ ๐™ข๐™š๐™ฃ๐™ฉ๐™ค๐™ง๐™จ๐™๐™ž๐™ฅ..

30/03/2025

๐‘พ๐’†๐’๐’„๐’๐’Ž๐’† ๐’•๐’ ๐‘ช๐‘บ๐‘ฐ๐‘น ๐‘ต๐‘ฌ๐‘ป ๐‘ณ๐’Š๐’‡๐’† ๐‘บ๐’„๐’Š๐’†๐’๐’„๐’† ๐’„๐’๐’‚๐’„๐’‰๐’Š๐’๐’ˆ, ๐’•๐’‰๐’† ๐’–๐’๐’•๐’Š๐’Ž๐’‚๐’•๐’† ๐’…๐’†๐’”๐’•๐’Š๐’๐’‚๐’•๐’Š๐’๐’ ๐’‡๐’๐’“ ๐‘ช๐‘บ๐‘ฐ๐‘น ๐‘ต๐‘ฌ๐‘ป ๐‘ณ๐’Š๐’‡๐’† ๐‘บ๐’„๐’Š๐’†๐’๐’„๐’† ๐’‚๐’”๐’‘๐’Š๐’“๐’‚๐’๐’•๐’”! ๐‘พ๐’† ๐’‘๐’“๐’๐’—๐’Š๐’…๐’† ๐’†๐’™๐’‘๐’†๐’“๐’• ๐’„๐’๐’‚๐’„๐’‰๐’Š๐’๐’ˆ, ๐’Š๐’-๐’…๐’†๐’‘๐’•๐’‰ ๐’”๐’•๐’–๐’…๐’š ๐’Ž๐’‚๐’•๐’†๐’“๐’Š๐’‚๐’๐’”, ๐’‚๐’๐’… ๐’”๐’•๐’“๐’‚๐’•๐’†๐’ˆ๐’Š๐’„ ๐’ˆ๐’–๐’Š๐’…๐’‚๐’๐’„๐’† ๐’•๐’ ๐’‰๐’†๐’๐’‘ ๐’š๐’๐’– ๐’†๐’™๐’„๐’†๐’ ๐’Š๐’ ๐‘ช๐‘บ๐‘ฐ๐‘น ๐‘ต๐‘ฌ๐‘ป ๐’‚๐’๐’… ๐’”๐’†๐’„๐’–๐’“๐’† ๐’‚ ๐’ƒ๐’“๐’Š๐’ˆ๐’‰๐’• ๐’‡๐’–๐’•๐’–๐’“๐’† ๐’Š๐’ ๐’“๐’†๐’”๐’†๐’‚๐’“๐’„๐’‰ ๐’‚๐’๐’… ๐’‚๐’„๐’‚๐’…๐’†๐’Ž๐’Š๐’„๐’”.

๐‘ฑ๐’๐’Š๐’ ๐’–๐’” ๐’‡๐’๐’“:
โœ… ๐‘บ๐’•๐’“๐’–๐’„๐’•๐’–๐’“๐’†๐’… ๐’„๐’๐’–๐’“๐’”๐’†๐’” ๐’…๐’†๐’”๐’Š๐’ˆ๐’๐’†๐’… ๐’ƒ๐’š ๐’”๐’–๐’ƒ๐’‹๐’†๐’„๐’• ๐’†๐’™๐’‘๐’†๐’“๐’•๐’”
โœ… ๐‘ช๐’๐’Ž๐’‘๐’“๐’†๐’‰๐’†๐’๐’”๐’Š๐’—๐’† ๐’๐’๐’•๐’†๐’”, ๐’Ž๐’๐’„๐’Œ ๐’•๐’†๐’”๐’•๐’”, ๐’‚๐’๐’… ๐’…๐’๐’–๐’ƒ๐’•-๐’„๐’๐’†๐’‚๐’“๐’Š๐’๐’ˆ ๐’”๐’†๐’”๐’”๐’Š๐’๐’๐’”
โœ… ๐‘ท๐’†๐’“๐’”๐’๐’๐’‚๐’๐’Š๐’›๐’†๐’… ๐’Ž๐’†๐’๐’•๐’๐’“๐’”๐’‰๐’Š๐’‘ ๐’•๐’ ๐’ƒ๐’๐’๐’”๐’• ๐’š๐’๐’–๐’“ ๐’„๐’๐’๐’‡๐’Š๐’…๐’†๐’๐’„๐’† ๐’‚๐’๐’… ๐’‘๐’†๐’“๐’‡๐’๐’“๐’Ž๐’‚๐’๐’„๐’†

๐‘บ๐’•๐’‚๐’š ๐’–๐’‘๐’…๐’‚๐’•๐’†๐’… ๐’˜๐’Š๐’•๐’‰ ๐’†๐’™๐’‚๐’Ž ๐’๐’๐’•๐’Š๐’‡๐’Š๐’„๐’‚๐’•๐’Š๐’๐’๐’”, ๐’”๐’•๐’–๐’…๐’š ๐’•๐’Š๐’‘๐’”, ๐’‚๐’๐’… ๐’”๐’–๐’„๐’„๐’†๐’”๐’” ๐’”๐’•๐’“๐’‚๐’•๐’†๐’ˆ๐’Š๐’†๐’”. ๐‘ณ๐’†๐’•'๐’” ๐’‚๐’„๐’‰๐’Š๐’†๐’—๐’† ๐’š๐’๐’–๐’“ ๐‘ช๐‘บ๐‘ฐ๐‘น ๐‘ต๐‘ฌ๐‘ป ๐’…๐’“๐’†๐’‚๐’Ž ๐’•๐’๐’ˆ๐’†๐’•๐’‰๐’†๐’“!

๐Ÿ“ฉ ๐‘ช๐’๐’๐’•๐’‚๐’„๐’• ๐’‡๐’๐’“ ๐’…๐’†๐’•๐’‚๐’Š๐’๐’”- +9160093 48115

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