Question

1)

A graph of the Absorbance vs Concentration of BSA at 545 nm has a linear trendline with the equation y = 0.6587x + 0.0001.

Please calculate the concentration of an unknown sample with an absorbance of 0.330.

		0.0001 mg/mL
		0.217 mg/mL
		0.6587 mg/mL
		0.501 mg/mL

2)

Please review the “Absorbance Assay at 280 nm” portion of your lab. Please calculate the concentration of an unknown sample with an absorbance of 0.330, path length 1.00 cm, and a molar absorption coefficient of 43824 M^–1cm^–1.  Please convert this from M to mg/mL for your answer.

Hint: Convert the molar concentration (M or moles/liter) of BSA from Absorbance Assay at 280 nm method to mg/mL, assume that the Molecular Weight of BSA = 66.5 kDa = 66500 g/mol

Answer 1

Q1. Correct answer is : 0.501 mg/mL

Q2. concentration of an unknown sample = 0.501 mg/mL

Explanation

Q1. The equation for linear trendline is : y = 0.6587x + 0.0001

Absorbance = 0.330 = y

From the equation,

x = (y - 0.0001) / 0.6587

x = (0.330 - 0.0001) / 0.6587

x = 0.501

concentration of BSA = 0.501 mg/mL

Q2. Absorbance = 0.330

According to Beer's law,

Concentration = (absorbance) / [(molar absorptivity) * (path length)]

Concentration = (0.330) / [(43824 M^-1cm^-1) * (1.00 cm)]

Concentration = 7.53 x 10^-6 M

Concentration = 7.53 x 10^-6 mol/L * (66500 g/mol) * (1 L / 1000 mL) * (1000 mg / 1 g)

Concentration = 0.50075 mg/mL