Category: CSIR NET Life Science Previous Year Questions and Solutions Spectroscopy

CSIR NET Life Science Previous Year Questions and Solutions – Spectroscopy

If you are preparing for the CSIR NET Life Science exam, mastering Spectroscopy is essential. This topic frequently appears in Part B and Part C questions, often testing both conceptual understanding and application-based problem solving. Practicing CSIR NET Life Science Previous Year Questions (PYQs) from Spectroscopy can significantly boost your preparation and help you score higher.

In this guide, you will find:

Importance of Spectroscopy in CSIR NET Life Science

Weightage in previous years

Solved Previous Year Questions from Spectroscopy

Preparation tips and study resources

Why Spectroscopy is Important in CSIR NET Life Science?
Spectroscopy forms the backbone of experimental biology and biochemistry. From studying protein structures to analyzing DNA, spectroscopic techniques such as UV-Vis, IR, NMR, Fluorescence, and Circular Dichroism are crucial.

In CSIR NET Life Science, questions on Spectroscopy often test:

Principles of different spectroscopic techniques

Absorbance and Beer–Lambert Law

Instrumentation and detectors

Applications in life sciences

Interpretation of spectra

Weightage of Spectroscopy in CSIR NET Life Science

Based on the analysis of last 10 years of CSIR NET Life Science PYQs, Spectroscopy appears almost every year:

Year No. of Questions Marks
2024 4 16
2023 3 12
2022 5 20
2021 4 16
2020 3 12

Trend: On average, 3–5 questions (12–20 marks) are directly from Spectroscopy, making it a high-return topic.

CSIR NET Life Science Previous Year Questions – Spectroscopy (With Solutions)
Question 1 – CSIR NET 2023 (Part B)
Which of the following statements regarding Beer–Lambert Law is correct?

A. Absorbance is inversely proportional to concentration.
B. Absorbance is directly proportional to path length.
C. Transmittance is directly proportional to concentration.
D. Wavelength has no effect on absorbance.

Answer: B
Solution: Beer–Lambert Law states:

𝐴
=
𝜀
𝑐
𝑙
A=εcl
where
𝐴
A = absorbance,
𝜀
ε = molar extinction coefficient,
𝑐
c = concentration,
𝑙
l = path length. Absorbance is directly proportional to both concentration and path length.

Question 2 – CSIR NET 2022 (Part C)
In a UV-Vis spectroscopy experiment, a solution shows maximum absorbance at 280 nm. What does this indicate about the sample?

Answer:
Absorbance at 280 nm is characteristic of aromatic amino acids like tryptophan and tyrosine in proteins. This indicates the presence of protein in the sample.

Question 3 – CSIR NET 2021 (Part C)
A protein sample shows a circular dichroism (CD) spectrum with a negative band at 222 nm and 208 nm. What is the predominant secondary structure?

Answer:
Negative bands at 222 nm and 208 nm in CD spectroscopy indicate an α-helical secondary structure.

Question 4 – CSIR NET 2020 (Part B)
Match the spectroscopic technique with its application:

Technique Application
NMR A. Protein secondary structure
IR B. Functional group identification
CD C. 3D protein structure determination

Answer:

NMR – C

IR – B

CD – A

Preparation Tips for Spectroscopy in CSIR NET Life Science
Understand the Principles: Don’t just memorize formulas – understand how each technique works.

Practice Numerical Problems: Many PYQs involve calculations using Beer–Lambert Law or energy-wavelength relationships.

Focus on Applications: Know which technique is used for which type of biological analysis.

Analyze PYQs: Identify question patterns and recurring concepts.

Revise with Summary Notes: Keep a quick revision sheet for all spectroscopy techniques.

Recommended Study Resources
CSIR NET Life Science Previous Year Question Papers – Practice all Spectroscopy-related questions.

Molecular Biology & Biophysics books – e.g., Principles of Instrumental Analysis by Skoog and Biophysical Chemistry by Upadhyay.

Let’s Talk Academy Online Coaching – Learn Spectroscopy concepts from experts like Suraj Sir, known as the God of Life Science.

Conclusion
Spectroscopy is a scoring section in the CSIR NET Life Science exam. By solving previous year questions with solutions, you not only strengthen your conceptual understanding but also gain the speed and accuracy needed in the exam.

If you want to master Spectroscopy for CSIR NET Life Science, make PYQ practice a daily habit. Remember, concept + practice = guaranteed marks.

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55. The molecular ion peak [M]⁺ of an analyte as measured by Electron Ionization Mass Spectrometry has an m/z of 149 and a relative abundance of 100%. The [M]⁺ has a relative abundance of 6.7% and the [M + 2]⁺ peak has a relative abundance of 5%. The abundance of the major isotope of H, C, N, O, and S are ¹H–100%, ¹²C–98.9%, ¹³C–1.1%, ¹⁴N–99.6%, ¹⁵N–0.4%, ¹⁶O–99.8%, ¹⁸O–0.2%, ³²S–95.0%, ³³S–0.75% and ³⁴S–4.2%. The most probable molecular formula of the compound is: (A) C₇H₂₁N₂O (B) C₅H₁₁NO₂S (C) C₆H₁₃O₂S (D) C₆H₁₅NOS

Determining Molecular Formula

54. For a nuclear spin of spin quantum number (I = 1/2), precessing in a magnetic field at a Larmor frequency of 300 MHz, the wavelength of incident radiation required to excite the nuclear spins must be approximately (A) 1 nm (B) 1 cm (C) 1 m (D) 10 m

Calculating the Wavelength of Radiation Required

53. Which one of the following correctly describes the spectroscopic experiment that would help distinguish between an α-helix, 3₁₀ helix and π helix? (A) Near UV absorption spectrum between 250–300 nm. (B) Fluorescence emission spectra between 350–400 nm. (C) ¹H NMR spectroscopy involving Hydrogen/Deuterium exchange. (D) Near UV Circular Dichroism spectrum between 250–300 nm.

Distinguishing α-Helix, 3₁₀ Helix, and π-Helix

52. The structure of a protein with 100 residues was determined by X-ray analysis at atomic resolution and NMR spectroscopy. The following observations are possible. A. The dihedral angles determined from the X-ray structure and NMR will be identical. B. The dihedral angles determined from the X-ray structure will be more accurate. C. β-turns can be determined only by NMR. D. β-sheets can be more accurately determined from the X-ray structure. Indicate the combination with ALL correct answers. (A) A and C (B) B and D (C) B and C (D) A and D

Comparison of X-ray Crystallography and NMR Spectroscopy

Spectroscopic Investigation of Proteins

50. Detailed NMR spectra of a 20-residue peptide were recorded using a 600 MHz instrument. If the peptide adopts an α-helical conformation, which one of the following statements is correct? (A) Prominent NHᵢ − NHᵢ₊₁ NOE peaks would be observed along with ³JNH–Hα coupling constants 8.5 Hz. (B) Prominent CαHᵢ − NHᵢ₊₁ NOE peaks would be observed along with ³JNH–Hα coupling constants 4.8 Hz. (C) Prominent CαHᵢ − NHᵢ₊₁ NOE peaks with ³JNH–Hα coupling constants 8.5 Hz. (D) Prominent NHᵢ − NHᵢ₊₁ NOE peaks along with ³JNH–Hα coupling constants 4.8 Hz.

Identifying Alpha-Helical Peptides Using NOE Patterns

49. The conformation of a 30-residue peptide is studied by NMR spectroscopy. The JNH for most of the amide protons is 4 Hz. The 2D NOESY spectrum shows prominent Nᵢ–Nᵢ₊₁ connectivities. The conformation of the peptide is (A) Anti-parallel β sheet. (B) Parallel β sheet. (C) Helix-like. (D) Unordered.

Identifying Helical Peptide Conformation

48. The hydrogen atoms in the δ (delta) methylene group of lysine will give the following splitting pattern in the ¹H-NMR spectra of lysine (A) Triplet of triplets. (B) Quintet. (C) Doublet of triplets. (D) Triplet of a doublet.

Splitting Pattern of the δ-Methylene Protons

47. The Nuclear Magnetic Resonance (1D and 2D) spectrum of a 30-residue peptide were recorded at 25°C. The following observations were made. A. The NH and CαH resonances were well resolved. B. The NOESY spectra showed extensive Nᵢ − Nᵢ₊₁ connectivities. C. The NH resonances showed slow exchange with deuterium. The spectra indicates that the peptide adopts (A) Helical conformations (B) Anti-parallel β-strand conformations (C) Polyproline conformation (D) β-turn conformation with four amino acids participating in the turn. Rest of the amino acids are unstructured.

Interpreting 1D and 2D NMR Spectra

46. A 30-residue peptide containing Phe, Tyr and Trp is dissolved in D₂O and the high field proton NMR is recorded after 24 hours. The resonances that are unlikely to be present are (A) Aromatic protons (B) Cα protons (C) Aliphatic protons (D) Amide protons

Effect of D₂O on Proton NMR of Peptides

$45. Point group symmetry operations such as inversion and mirror plane are not applicable to protein crystals. This is because (A) Protein molecules assemble in highly ordered fashion. (B) Protein molecules have handedness. (C) Protein molecules form a lattice plane that do not diffract X-rays. (D) Hydrogen atoms in proteins diffract weakly.$

Why Mirror Plane and Inversion Symmetry

Protein Conformational Dynamics

$43. From the following statements, A. Surface plasmon resonance can be used to determine binding constants only in the range of 10²–10³ M. B. De novo sequencing is not possible by mass spectral methods. C. The position of hydrogen atoms in proteins is not directly determined by X-ray diffraction. D. Circular dichroism and nuclear magnetic resonance spectroscopy do not give the same information on protein structure. Choose the option with all correct statements. (A) A, B, C (B) A, C, D (C) B, D (D) C, D$

Protein Structure Analysis Techniques

42. Shown below are the CD spectra of a protein recorded under two different conditions. From the options given below, select the one that is the best interpretation of the spectra. (A) The protein has a helical secondary structure under condition A that is denatured under condition B. (B) The protein has a helical secondary structure under condition A that is converted to β sheets under condition B. (C) The spectra represent the tertiary fold of the protein with condition A corresponding to mixed α helix + β sheet fold and condition B corresponding to largely β sheet fold. (D) The difference between the spectra under conditions A and B is due to lower protein concentration under condition B.

Interpreting Far-UV Circular Dichroism Spectra

41. Protein 'A' was subjected to different experiments: (i) SDS-PAGE with/without β-mercaptoethanol (β-ME) (ii) Fluorescence (iii) Far-UV CD (iv) Near-UV CD spectra at pH 7.0 (black) and 3.0 (red) The results are shown below. Which one of the following options provides the correct inference? (A) Protein 'A' is an S-S bonded homotetramer and each subunit has a molecular mass of 50 kDa, folded at pH 7.0 and molten globule at pH 3.0. (B) Protein 'A' has a molecular mass of 200 kDa, β-ME degrades the protein, low pH changes the conformation from α-helix to β-sheet. (C) SDS denatures protein 'A' into different oligomeric states, low pH changes the conformation from α-helix to β-sheet. (D) SDS promotes the formation of different oligomeric states of Protein 'A', low pH changes the conformation from β-sheet to α-helix.

Interpreting SDS-PAGE, Fluorescence

40. A circular dichroism spectrum in the far-UV region informs on the kind and content of secondary structures in a protein. Near-UV and tryptophan emission spectra inform on the tertiary structure. Shown in the panels above are (A) intrinsic fluorescence emission spectra of protein 'X', (B) far-UV CD spectra of protein 'X', (C) near-UV CD spectra of protein 'X' recorded under different conditions. Curves represent the spectra of protein X at pH 7.0 (black), pH 3.0 (green), and pH 7.0 in the presence of 6.0 M guanidine hydrochloride (red). What does the experiment report? (A) Protein is fully folded at pH 7.0, acid-induced molten globule at pH 3.0, and unfolded in 6 M guanidine hydrochloride. (B) Protein secondary structure is reduced at pH 7.0 and the protein has formed β-fibrils at the other two conditions. (C) The changes in fluorescence and near-UV CD indicate increase in hydrodynamic radius at pH 3.0 and in 6 M guanidine hydrochloride. (D) There is extensive denaturation of the protein both at pH 3.0 and in 6 M guanidine hydrochloride.

Interpreting Fluorescence, Far-UV CD, and Near-UV CD Spectra

39. A researcher is investigating structural changes in a protein by following tryptophan fluorescence and circular dichroism. Fluorescence and CD spectra of the pure protein were obtained in the absence of any treatment (A), in the presence of 0.5 M urea (B), upon adding acrylamide, a quencher of tryptophan (C), and upon heating (D). The data are shown below. Which one of the following statements is correct? (A) CD is more sensitive to structural changes than fluorescence. (B) Fluorescence is more sensitive to structural changes than CD. (C) Both the methods are equally responsive to structural changes. (D) Acrylamide alters the secondary structure of the protein.

Fluorescence vs Circular Dichroism

38. A 100 residue long protein has a single chromophoric residue (tyrosine). The UV absorption of this protein and a homologous protein (also with a single tyrosine residue) was monitored at 280 nm at different pH conditions. A plot of the absorbance as a function of pH is shown below. The locations of the tyrosine residue in the context of the protein sequence is also shown in the figure. Which one of the following rationalises the difference in the two pH titrations? (A) Removal of the hydroxyl group of tyrosine above pH 11. (B) Location of the tyrosine residue in the protein structure. (C) pH dependent changes in the absorption in the polypeptide main chain. (D) Hydrolysis of the polypeptide as a function of pH.

Effect of Tyrosine Environment on UV Absorption

37. Cytochrome-c has only one tryptophan residue (W) which is buried. The protein in cacodylate buffer (pH 6.0) is excited at 280 nm, and its emission spectrum measured in the range of 300–450 nm. The same measurement was repeated on the protein in the buffer containing 6 M guanidine hydrochloride. It was observed that there is an increase in the intensity of the emission spectrum of the guanidine hydrochloride-treated cytochrome-c. The most probable reason for this increase is: (A) W is near a hydrophobic patch present in the unfolded protein. (B) W is near heme in the native protein. (C) W is near carboxylate amino acid side chains in the native protein. (D) W is in a polar pocket in the native protein.

Fluorescence Emission of Cytochrome-c

36. The mechanism of oxygen transport by hemocyanin (containing Cu) is described by: Cu⁺ + Cu⁺ + O₂ ⇌ Cu²⁺–O₂²⁻–Cu²⁺ Which one of the following techniques can be used to monitor the change in the oxidation state of copper? (A) Mass spectrometry (B) Circular dichroism (C) Absorption spectroscopy (D) Fluorescence spectroscopy

Monitoring Copper Oxidation State

35. Which of the following atomic nuclei cannot be probed by nuclear magnetic resonance spectroscopy? (A) ¹H (B) ³¹P (C) ¹⁸O (D) ¹⁵N

Nuclei That Can and Cannot Be Studied

34. Haemoglobin has characteristic circular dichroism (CD) peaks in the far-UV, near UV and Soret regions. Contribution to near-UV CD comes entirely from (A) Aromatic amino acid residues. (B) Heme group. (C) Heme and aromatic amino acid residues. (D) Peptide bonds and aromatic amino acid residues.

Near-UV Circular Dichroism of Hemoglobin

33. Following observations are made regarding a peptide sequence. - The peptide is inert to Ellman's reagent. However, on reacting with β-mercaptoethanol, the peptide gives a positive Ellman's test. - The peptide sequence gives a broad minimum around 211 nm in the CD spectrum. - With increasing concentration of the peptide, the melting temperature of the peptide increases. - On treating the peptide with D₂O, half the total number of amides get exchanged. Which one of the following statements is correct? (A) It is an α-helical peptide that undergoes aggregation. (B) It is an α-helical disulfide-bridged peptide that undergoes aggregation. (C) It is a β-hairpin peptide, which is stabilized by a disulfide bridge. (D) The peptide is composed of an α-helix and β-sheet connected by a disulfide bridge.

Identifying a Disulfide-Bridged β-Hairpin Peptide

32. Poly-L-lysine exists in pure α-helix, β-sheet and random coiled conformation depending upon the solvent conditions. The values of mean residue ellipticity at 220 nm ([Θ]₂₂₀) are −35,700, −13,800 and +3,900 deg cm² dmol⁻¹ for α-helical, β-sheet and random coil conformations of this polypeptide, respectively. The polypeptide exists in α-helix conformation at pH 10.8 and 25°C. Addition of urea leads to a two-state transition between α-helix and random coil conformation. It has been observed that [Θ]₂₂₂ of the polypeptide is −14,800 deg cm² dmol⁻¹ in the presence of 6 M urea. The percentage of the polypeptide in α-helix conformation is: (A) 37 (B) 41 (C) 47 (D) 50

Calculating Alpha-Helix Percentage

$31. A polymer is synthesized from an achiral amino acid. Conformation of the polymer can be investigated by the following techniques. A. Fibre diffraction B. Nuclear magnetic resonance spectroscopy C. Circular dichroism spectroscopy D. Differential scanning calorimetry Choose the combination which would indicate that the polymer adopts a helical conformation. (A) A, C (B) B, D (C) A, B (D) C, D$

Identifying Helical Conformation of a Polymer

30. At 25°C values of [Θ]₂₂₂, the mean residue ellipticity at 222 nm, are −33,000 and −3,000 deg cm² dmol⁻¹ for a polypeptide existing in α-helical (α) and β-structure (β), respectively. If this polypeptide undergoes a two-state heat-induced α → β transition, and a value of [Θ]₂₂₂ = −18,000 deg cm² dmol⁻¹ is observed at 60°C, then this observation leads to the conclusion that the α helix conversion to β-structure: (A) 40% (B) 50% (C) 55% (D) 60%

Alpha-Helix to Beta-Sheet Conversion

$29. A protein polypeptide chain exists in α-helical conformation in a solvent and it has a value of −30,000 deg cm² dmol⁻¹ for the mean residue ellipticity at 222 nm ([Θ]₂₂₂) in the temperature range 20–50°C. On raising the temperature above 50°C, [Θ]₂₂₂ increases and reaches a value of −2,000 deg cm² dmol⁻¹ at 70°C and the value of [Θ]₂₂₂ remains unchanged above 70°C. The observed value of [Θ]₂₂₂ is −14,000 deg cm² dmol⁻¹ at 60°C. If one assumes that the heat-induced denaturation is a two-state process, the fraction of α-helix at 60°C is (A) 0.40 (B) 0.43 (C) 0.50 (D) 0.57$

Fraction of Alpha-Helix from Circular Dichroism Ellipticity

28. A 26-residue peptide composed of alanine and leucine shows a circular dichroism (CD) spectrum characteristic of α-helix at 50°C in 5 mM phosphate buffer at pH 7.4. Deconvolution of the spectrum indicates 60% α-helical and 40% random conformation. When the peptide solution is cooled gradually to 25°C, and the CD spectra are recorded at different temperatures, the most likely observation will be that (A) The % helical content will decrease and % random conformation will increase. (B) The % helical content will increase and % random conformation will decrease. (C) There will be transition from α-helix to β-sheet. (D) There will be transition from α-helix to β-hairpin.

Alpha-Helix Stability During Cooling in Circular Dichroism

27. Which one of the following statements is correct? (1) Electrospray ionization mass spectrum of a compound can be obtained only if it has a net positive charge at pH 7.4 (2) Helical content of a tryptophan containing peptide can be obtained by examining the fluorescence spectrum of tryptophan (3) The occurrence of beta sheet in a protein can be inferred from its circular dichroism spectrum (4) The chemical shift spread for a compound is more in its 1H NMR spectrum as compared to its 13C NMR spectrum

How Circular Dichroism Identifies Beta-Sheet Structure

26. Two homologous proteins were isolated from a psychrophile (P) and a thermophile (T). The purified proteins were subjected to denaturation, protease digestion and circular dichroism (CD). Following observations were made: A. The CD spectra of P and T proteins are identical B. Their amino acid composition is 95% identical C. T and P are equally susceptible to proteolysis in the presence or absence of reducing agent D. T has higher midpoint of thermal denaturation than P The reason for enhanced stability in T is due to (1) Altered secondary structure (2) Increased number of disulfides in T (3) Increase in water of hydration (4) Increase in number of salt bridges

Why Are Thermophilic Proteins More Stable

$25. The structure of a protein is known from X-ray diffraction studies which gave 30% α-helix, 50% β-sheet and 20% random coil. Circular dichroism (CD) measurements gave 50% α-helix, 40% β-sheet and 10% random coil. What could NOT be a possible explanation for these observations. (MP) (1) Protein structure in the crystal is different from that in the solution. (2) CD analysis for structural components is not appropriate for this protein. (3) Contributions from other chromophores also contribute to the CD spectrum of the protein. (4) Protein contains high content of disulphide bonds.$