Question

1. Max needs to estimate the protein concentration of a cell lysate. He first read the absorbance of 100 uL aliquots of his sample at 280 nm and calculated an average absorbance of 0.623. Next, using this information he plans to dilute his sample for a Lowry assay.

a) Based upon the A280 readings, what was the estimated concentration of Max’s cell lysate?

b) Max wants to make 1:2, 1:4, 1:10, and 1:20 dilutions of his cell lysate. Describe what volumes of his cell lysate and water would be needed to prepare 50 uL of each of these dilutions.

c) Using the estimation from part a), determine the estimated protein concentrations of each of these dilutions and determine the amount of protein Max should expect to have if he uses 5 uL aliquots in his Lowry assay

Answer 1

At 280 nm, aromatic amino acids such as tyrosine and tryptophan and also cysteine residues absorb at this wavelength. For the determination of protein concentrations at 280 nm in cell lysates which have a mixture of proteins, the basic formula that can be used is :

Protein concentration (mg / mL) = Absorbance at 280 nm / path length (cm)

Here, the volume used in the cuvette is not considered for calculation, as long as you are able to use a volume that gives an absorbance, you can convert the value into mg / mL protein.

The path length for most spectrophotometric measurements is usually 1 cm, so the concentration of protein would simply be the absorbance value in mg / mL.

a) Given the average absorbance value to be 0.623. Using the above formula, we can arrive at the concentation of protein in the cell lysate to be: 0.623 / 1 = 0.623 mg / mL.

b) The dilutions given are 1:2, 1:4, 1:10, and 1:20, they can be better written as 1 / 2 , 1 / 4, 1 / 10 and 1 / 20. These would mean that the sample is diluted 2,4,10 and 20 times respectively. The total volume for making each dilution is 50 uL.

The sample volume required for each dilution can be obtained by multiplying the required dilution by the tota volume:

For a 1/2 dilution, the sample volume would be: 1 / 2 * 50 = 0.5 * 50 = 25 uL

So, we need a 25 uL sample in 50 uL to get a 1 / 2 dilution: 25 / 50 = 1 / 2.

The volume of lysate here is 25 uL and the volume of water would be 25 uL ( 50 - 25 = 25).

Similarly, for the other dilutions:

1 / 4 dilution: 1 / 4 * 50 = 0.25 * 50 = 12.5 uL lysate

Volume of water = 50 - 12.5 = 37.5 uL.

1 / 10 dilution: 1 / 10 * 50 = 0.1 * 50 = 5.0 uL lysate.

Volume of water = 50 - 5 = 45 uL.

1 / 20 dilution: 1 / 20 * 50 = 0.05 * 50 = 2.5 uL lysate.

Volume of water = 50 - 2.5 = 47.5 uL.

A more simplified way of looking at this is 1 / 2 means we are diluting the sample in to half its original concentration so we take the sample and add equal volume of water. Here we fix the total volume as 50 uL, for a 1 / 2 dilution, we are taking 25 uL lysate which is half the volume of 50 uL. 1 /4 is double the dilution of 1 / 2 , and so we are taking half the volume (12.5 uL) of the lysate we took for 1 / 2 . A 1 / 10 dilution means we need to take one-tenth of the original sample to make a dilution. one-tenth of 50 u would be 5 uL. A 1 / 20 is double the dilution of 1 / 10 so we take half the volume (2.5 uL) of sample we took for 1 /10 to make a 1 /20.

c) Now, we are asked to determine the protein concentrations of the dilutions prepared and the amount of protein Max would expect to get if he took 5uL of each of these dilutions for the Lowry assay.

We calculated the original concentration of the sample to be : 0.623 mg / mL

Now, for each of the dilutions, we would determine the concentrations by multiplying the above value with each of the dilutions:

Undiluted lysate: 0.623 mg / mL

1 / 2 dilution : 0.623 * 1 / 2 = 0.623 * 0.5 = 0.31 mg / mL.

1 / 4 dilution: 0.623 * 1/4 = 0.623 * 0.25 = 0.16 mg / mL

1 / 10 dilution: 0.623 * 1/ 10 = 0.623 * 0.1 = 0.0623 mg / mL

1 / 20 dilution: 0.623 * 1 / 20 = 0.623 * 0.05 = 0.031 mg / mL

Now, Max uses 5 uL each from the above dilutions, so we need to determine the amount of protein used by multiplying each of the values by 0.005 ( 5 uL = 0.005 mL) to get the protein values:

1/2 dilution: 0.31 * 0.005 = 0.0016 mg or 1.6 ug protein( 1 mg = 1000 ug)

1/4 dilution: 0.16 * 0.005 = 0.0008 mg or 0.8 ug protein

1 /10 dilution: 0.0623 * 0.005 = 0.00031 mg or 0.3 ug protein

1 / 20 dilution: 0.031 * 0.005 = 0.00016 mg or 0.16 ug protein.

Lowry's method is sensitive between 5 and 100 ug protein and so none of these samples would give results based on the calculated values unless the sample volume of 5 uL for estimation is considerably increased.