Q.35 If a 10 mM solution of a biomolecule in a cuvette of path length 10 mm absorbs
90% of the incident light at 280 nm, the molar extinction coefficient of the
biomolecule at this wavelength is _____________ M-1cm-1.
(Round off to two decimal places)
The molar extinction coefficient is 100.00 M⁻¹ cm⁻¹. This value is calculated using the Beer-Lambert law from the given absorbance, concentration, and path length. The detailed solution follows standard spectrophotometry principles relevant for CSIR NET Life Sciences preparation.
Problem Breakdown
A 10 mM biomolecule solution absorbs 90% of light at 280 nm in a 10 mm path length cuvette.
Therefore, A = −log₁₀(0.10) = 1.00
- Concentration c = 10 mM = 0.010 M
- Path length l = 10 mm = 1.00 cm
Beer-Lambert Law Application
The law states: A = ε · c · l, rearranged to ε = A/(c · l)
This matches protein UV absorbance calculations at 280 nm due to aromatic residues (Trp, Tyr).
Calculation Verification
The query specifies a numerical answer (no multiple-choice options). Rounding to two decimal places gives exactly 100.00, as −log₁₀(0.10) = 1 precisely.
Key Steps in Calculation
- Transmittance to Absorbance: 90% absorption means 10% transmittance (T=0.10), so A=−log₁₀(0.10)=1.00
- Units Conversion: c=10×10⁻³ M; l=1 cm
- Formula Application: ε=1.00/(0.010×1)=100.00 M⁻¹ cm⁻¹
Practical Relevance for CSIR NET
This tests spectrophotometry fundamentals in biochemistry and biophysics. Typical protein ε₂₈₀ ranges 1000–50,000 M⁻¹ cm⁻¹, but here it’s derived experimentally. Use for protein quantification without standards.
Common Pitfalls to Avoid
- Forgetting path length conversion (cm, not mm)
- Miscalculating absorbance from percentage absorption
- Unit mismatch between mM and M concentrations
