You are given a pure protein sample to characterize and provided the following i

Question

You are given a pure protein sample to characterize and provided the following information: Its molar extinction coefficient, epsilon280, is 0.30 liters micromole-1 cm-1 Its DeltaGdegree for unfolding is 4.0 kcal/mol at 37degree (where RT = 0.59 kcal/mole) Using a 1 cm path length cell, you measure the absorbance at 280 nm of a 15-fold dilution of your pure protein in solution and find A280 = 0.60. What is the original concentration of the protein before dilution? What is the concentration of the unfolded form of the protein in your sample?

Explanation / Answer

a) Use Beer's Law:

A = lc

A = 0.60

= 0.3 L/mol*cm

l = 1 cm

Solve for c = A/(l) = 0.60/(0.3*1) = 2 mol/L

Since it was a 15 fold dilution, the pre-dilution concentration is 30 mol/L or 30 M

b) The protein will undergo the following process:

folded ----> unfolded

G for this proces is 4.0 kcal/mol

Also,

G = -RT ln(K), where K is the equilibrium constant, K = [unfolded]/[folded] and RT = 0.59 kcal/mol

Solve for K:

K = exp(-G/RT) = [unfolded]/[folded]

Therefore:

[unfolded] = [folded]*exp(-G/RT) = 30 M * exp(-4/0.59) = 0.034 M or 34 nM

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