Question

A dairy claims that its skim milk has a protein concentration of 6000 mg in 250 ml of milk. You plot a standard curve using casein – a protein which constitutes 80% of the protein found in milk. Your plot yields a equation of the line: y = 12.6977x + 0.1370 where y is the absorbance and x is given in mg casein/ml solution. The dairy milk has an absorbance of 5.08 when a 1/60 dilution is analyzed. Do you agree with the dairy?

Answer 1

The claimed protein concentration of the skim milk is = 6000 mg / 250 mL = 24 mg/mL

Now, the from the equation we can find the experimental value of protein concentration in the skim milk as follows:

y = 12.6977x + 0.1370

or, x = (y - 0.1370)/ 12.6977

or, x = (5.08 -0.1370)/ 12.6977

or, x = 0.3893 mg/mL

but this protein concentration is at 1/60 dilution so we have to take the claimed concentration at 1/60 dilution as well.

Therefore, the claimed protein concentration at 1/60 dilution is = (24 mg/mL) / 60

= 0.4 mg/mL

Therefore, the experimental and claimed values of protein concentration are in good agreement, the small discrepancy arising is probably due to the fact that casein constitute 80% of the protein content and we have not considered the other 20% of the protein in our experiment.

So, we would agree with the dairy.