Q.42 A 0.1% (w/v) solution of a protein absorbs 20% of the incident light. What fraction of light is
transmitted if the concentration is increased to 0.4%? [Correct to two decimal places]

Protein solutions follow the Beer-Lambert law, where absorbance scales linearly with concentration, allowing calculation of transmitted light at higher concentrations. For a 0.1% (w/v) solution absorbing 20% of incident light, the transmitted fraction at 0.4% (w/v) is 0.41.​

Problem Breakdown

A 0.1% (w/v) protein solution absorbs 20% of incident light, meaning transmittance T1=1−0.20=0.80T1=1−0.20=0.80. Absorbance A1=−log⁡10(T1)=−log⁡10(0.80)≈0.097A1=−log10(T1)=−log10(0.80)≈0.097. Since concentration increases from 0.1% to 0.4% (4-fold), new absorbance A2=4×A1≈0.388A2=4×A1≈0.388.​

Calculation Steps

Transmittance relates to absorbance by T=10−AT=10−A. Thus, T2=10−0.388≈0.41T2=10−0.388≈0.41 (correct to two decimal places). This assumes constant path length and molar absorptivity, as per Beer-Lambert law $A=ϵlc$.​

Introduction to Protein Solution Absorbance Calculation

In biochemistry, protein solution absorbance measures light absorption to determine concentration via the Beer-Lambert law. A classic CSIR NET question asks: A 0.1% (w/v) solution of a protein absorbs 20% of the incident light. What fraction of light is transmitted if concentration is increased to 0.4%? [Correct to two decimal places]. This problem tests understanding of transmittance, absorbance, and concentration proportionality, key for biotech competitive exams.​

Beer-Lambert Law Explained

The Beer-Lambert law states $A=ϵlc$, where absorbance $A$ is proportional to concentration $c$. Transmittance T=I/I0=10−AT=I/I0=10−A, and absorption fraction = $1-T$. For the 0.1% protein solution absorbing 20%, T1=0.80T1=0.80, so A1=−log⁡10(0.80)≈0.097A1=−log10(0.80)≈0.097.​

• At 0.4% concentration, c2/c1=4c2/c1=4, so A2=4×0.097=0.388A2=4×0.097=0.388.​

• Transmitted fraction T2=10−0.388=0.41T2=10−0.388=0.41.​

Step-by-Step Solution for CSIR NET

1. Calculate initial transmittance: T1=1−0.20=0.80T1=1−0.20=0.80.​

2. Find A1=−log⁡10(0.80)≈0.0969A1=−log10(0.80)≈0.0969.​

3. Scale absorbance: A2=(0.4/0.1)×A1=0.3876A2=(0.4/0.1)×A1=0.3876.​

4. Compute T2=10−A2≈0.41T2=10−A2≈0.41.​

This matches exam solutions where fraction transmitted is 0.40 (rounded), but precise value is 0.41.​

Why No Options Provided?

The query mentions “explain every option,” but this numerical problem from IIT JAM/CSIR NET lacks MCQs. Common distractors include linear scaling of transmittance (wrong, yields 0.20) or absorption (1.00, impossible). Always use logarithmic relation for accuracy.​

Applications in Protein Quantification

UV absorbance at 280 nm quantifies proteins non-destructively. For 0.1% (w/v) solutions, extinction coefficients help convert $A$ to concentration. Ideal for enzyme kinetics and biotech research relevant to CSIR NET Life Sciences.​