Protein Molar Extinction Coefficient Calculator

Molar Extinction Coefficient (ε) The molar extinction coefficient ε (also called molar absorptivity) quantifies how strongly a protein absorbs light at 280 nm. Beer-Lambert law: A = ε × l × c Where A = absorbance, l = path length (cm), c = molar concentration (M). At 280 nm, absorption is dominated by aromatic amino acids.

Pace Formula (1995) ε₂₈₀ = (nTrp × 5,500) + (nTyr × 1,490) + (nCys-Cys × 125) Where: nTrp = number of tryptophan residues (absorbs most strongly) nTyr = number of tyrosine residues nCys-Cys = number of disulfide bonds (cystine), NOT free cysteine

Units: M⁻¹cm⁻¹ (also written as L·mol⁻¹·cm⁻¹) Source: Pace et al., Protein Science, 1995.

Predicting Protein Concentration From absorbance at 280 nm: c (mg/mL) = (A₂₈₀ × MW) / (ε₂₈₀ × path length) c (μM) = (A₂₈₀ × 10⁶) / (ε₂₈₀ × path length)

Why 280 nm? Trp and Tyr absorb UV light near 280 nm due to their aromatic rings. This is convenient because buffers, salts, and most contaminants don’t absorb at 280 nm. DNA and RNA absorb strongly at 260 nm, and their tail reaches 280, so nucleic acid contamination pushes A280 up and makes the protein look more concentrated than it is. The A260/A280 ratio is the standard check: clean protein sits near 0.6, clean DNA near 2.0. Phenylalanine (Phe) has very weak absorption at 280 nm; usually ignored.

Typical Values A protein with no Trp and no Tyr: ε ≈ 0, so A280 cannot be used at all Typical enzyme (2 to 5 Trp): ε ≈ 10,000 to 30,000 M⁻¹cm⁻¹ Antibodies: ε ≈ 200,000 to 220,000 M⁻¹cm⁻¹ BSA (bovine serum albumin, 66,463 Da): ε ≈ 43,824 M⁻¹cm⁻¹ Lysozyme: ε ≈ 36,000 M⁻¹cm⁻¹

Worked example

The placeholder values describe a 45.2 kDa protein carrying 2 tryptophans, 5 tyrosines and 1 disulfide bond.

ε₂₈₀ = (2 × 5,500) + (5 × 1,490) + (1 × 125) = 11,000 + 7,450 + 125 = 18,575 M⁻¹cm⁻¹

Read A280 = 0.850 in a 1 cm cuvette and the concentration follows straight from Beer-Lambert:

c = 0.850 ÷ (18,575 × 1) = 4.5760 × 10⁻⁵ M = 45.76 μM

Multiply by the molecular weight to get the number a bench protocol actually wants: 4.5760 × 10⁻⁵ mol/L × 45,200 g/mol = 2.068 mg/mL.

The useful shortcut falls out of the same two numbers. A 1 mg/mL solution of this protein reads ε ÷ MW = 18,575 ÷ 45,200 = 0.411 at 280 nm. Divide any A280 reading by 0.411 and you have mg/mL without touching the molar route at all. Every protein has its own version of that constant, and it is worth writing on the tube.

Notice how little the disulfide bond contributes: 125 out of 18,575, which is 0.7%. Miscounting cystines barely moves the answer. Miscounting tryptophans moves it a lot, since each one is worth 5,500.