10. Protein X has two tryptophans and one tyrosine, and its molecular weight is
20,000 Da. One molar tryptophan and one molar tyrosine solutions give you Abs280 of
5500 and 1490, respectively (using a one cm path length cuvette). What Abs280 would
you expect for 0.2 mg/ml of protein X solution?
a. 0.1249
b. 1.249
c. 12.49
d. This cannot be estimated from the given data
Protein X, with two tryptophans and one tyrosine, has a molar extinction coefficient at 280 nm of 12,490 M⁻¹ cm⁻¹, calculated as (2 × 5500) + (1 × 1490). For a 0.2 mg/ml solution and molecular weight of 20,000 Da, the molar concentration is 10 μM (0.2 / 20,000 × 1000). Thus, expected Abs₂₈₀ = 12,490 × 10⁻⁵ = 1.249 (path length 1 cm).
Correct Answer
b. 1.249. This matches Beer’s Law: A = εcl, where ε is the protein’s extinction coefficient, c is molar concentration, and l = 1 cm.
Option Analysis
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a. 0.1249: Too low; arises from mistakenly using 1 μM instead of 10 μM molar concentration (error in mg/ml to M conversion).
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b. 1.249: Correct; precise calculation using given ε values for Trp/Tyr and protein specifics.
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c. 12.49: Too high; assumes 1 mg/ml or 100 μM concentration, ignoring the 0.2 mg/ml value.
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d. This cannot be estimated: Incorrect; sufficient data (aromatic residues, MW, ε values, concentration) allows estimation via standard method.
Introduction to Protein Abs280 Calculation
Protein Abs280 calculation relies on UV absorbance at 280 nm from aromatic residues like tryptophan (Trp) and tyrosine (Tyr), key for protein X two tryptophans one tyrosine quantification in solutions like 0.2 mg/ml. Standard molar extinction coefficients (ε_Trp = 5500 M⁻¹ cm⁻¹, ε_Tyr = 1490 M⁻¹ cm⁻¹) enable precise Abs280 prediction for proteins of known composition and molecular weight (MW), such as 20,000 Da. This method suits CSIR NET Life Sciences exam prep, avoiding dyes for direct, rapid assays.
Step-by-Step Protein X Abs280 Solution
Calculate ε_protein = (2 × 5500) + (1 × 1490) = 11,000 + 1,490 = 12,490 M⁻¹ cm⁻¹. Molar concentration c = (0.2 mg/ml) / (20,000 g/mol) = 0.01 mg/ml ÷ 20 g/L equivalent = 10⁻⁵ M (10 μM). Abs₂₈₀ = ε × c × l = 12,490 × 10⁻⁵ × 1 = 1.249.
| Parameter | Value | Role in Abs280 |
|---|---|---|
| Trp residues | 2 | Major contributor (5500 each) |
| Tyr residues | 1 | Minor contributor (1490) |
| ε_total | 12,490 M⁻¹ cm⁻¹ | Protein-specific |
| MW | 20,000 Da | For c conversion |
| Mass conc. | 0.2 mg/ml | Query input |
| Expected Abs₂₈₀ | 1.249 | Final result |
Why Options Fail in Abs280 Estimation
Option a (0.1249) underestimates c by 10-fold; c (12.49) overestimates by 10-fold; d ignores viable data despite no cystine info needed here. Trp dominates (88% of ε), making two tryptophans one tyrosine setups predictable.
CSIR NET Applications
Protein absorbance 280 nm questions test Beer’s Law mastery for biotech assays, enzyme kinetics, and purification—vital for Life Sciences exams. Practice with tools like ProtParam confirms ε values.
