Q.16 Solutions of the following peptides are prepared separately at a concentration of 1 mM. Among
these four, which one has the highest A280?
(A) Ser-Val-Trp-Asp-Phe-Gly-Tyr-Trp-Ala
(B) Gln-Leu-Glu-Phe-Thr-Leu-Asp-Gly-Tyr
(C) Met-Gly-Val-Ileu-Asp-Ser-Ala-Trp-His
(D) His-Pro-Gly-Asp-Val-Ileu-Phe-Met-Leu
Option (A) has the highest A280 absorbance at 1 mM concentration. This occurs because A280 measures UV light absorption primarily from aromatic amino acids tryptophan (Trp) and tyrosine (Tyr), with Trp contributing the most due to its high molar extinction coefficient. All peptides have identical concentrations and similar lengths (9 residues), so the one with the most Trp-Tyr residues shows the strongest absorbance.
Amino Acids Absorbing at 280 nm
Tryptophan (Trp, W) has the strongest absorbance (ϵ≈5500−5690 M−1cm−1), followed by tyrosine (Tyr, Y; ϵ≈1200−1490). Phenylalanine (Phe, F) contributes minimally (ϵ≈200), while histidine (His, H), methionine (Met, M), and others do not significantly absorb. Peptide absorbance follows the additive rule: A280=ϵ×c×l (here, c=0.001 M, l=1 cm assumed), so higher ϵ yields higher A280.
Option Analysis
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(A) Ser-Val-Trp-Asp-Phe-Gly-Tyr-Trp-Ala: 2 Trp + 1 Tyr + 1 Phe. Highest Trp count drives maximum ϵ≈5500×2+1490×1+200×1=12690 M−1cm−1; A280≈12.7.
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(B) Gln-Leu-Glu-Phe-Thr-Leu-Asp-Gly-Tyr: 0 Trp + 1 Tyr + 1 Phe. ϵ≈1490+200=1690; A280≈1.7.
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(C) Met-Gly-Val-Ileu-Asp-Ser-Ala-Trp-His: 1 Trp + 0 Tyr + 0 Phe. ϵ≈5500; A280≈5.5.
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(D) His-Pro-Gly-Asp-Val-Ileu-Phe-Met-Leu: 0 Trp + 0 Tyr + 1 Phe. ϵ≈200; A280≈0.2.
Option A clearly dominates due to two Trp residues.
Peptide A280 absorbance determines protein/peptide concentration via UV spectroscopy at 280 nm, crucial for CSIR NET Life Sciences exams. This guide solves the exact question on four 1 mM peptide solutions, explaining aromatic amino acid roles for competitive exam success.
Why A280 Measures Peptides
A280 relies on aromatic residues: Trp (ϵ280=5500 M−1cm−1) dominates, Tyr adds 1490, Phe minor at 200. No Cys disulfides here, so ignore that term. For equal concentrations, count Trp > Tyr determines winner.
Detailed Option Breakdown
| Option | Sequence Highlights | Trp | Tyr | Phe | Approx. ϵ280 (M−1cm−1) | A280 (1 mM, 1 cm) |
|---|---|---|---|---|---|---|
| (A) | Ser-Val-Trp-Asp-Phe-Gly-Tyr-Trp-Ala | 2 | 1 | 1 | 12,690 | 12.7 |
| (B) | Gln-Leu-Glu-Phe-Thr-Leu-Asp-Gly-Tyr | 0 | 1 | 1 | 1,690 | 1.7 |
| (C) | Met-Gly-Val-Ileu-Asp-Ser-Ala-Trp-His | 1 | 0 | 0 | 5,500 | 5.5 |
| (D) | His-Pro-Gly-Asp-Val-Ileu-Phe-Met-Leu | 0 | 0 | 1 | 200 | 0.2 |
Answer: (A) – Two Trp residues give unmatched absorbance.
CSIR NET Exam Tips
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Prioritize Trp count in A280 questions.
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Use Edelhoch/Pace formula: ϵ=(nTrp×5500)+(nTyr×1490)+(nCys−Cys×125).
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Practice similar IIT JAM BT/CSIR queries for quick solving.
