Question

You analyzed the C18 content of a sample of ground beef. You took 5 g of beef, homogenized into buffer and centrifuged. The volume of the supernatant was 14 ml.

You took 2 ml, of the supernatant, extracted the fatty acids and then made the fatty acid methyl ester (FAME) derivatives.

The final volume of the FAME sample was 50 µl. 3 µl of this were injected onto the GC. The area of the C18 peak was 14753.

2µl of a 0.01 mg/ml C18 standard were injected in a second run. The area of the C18 peak was 18334.

What was the concentration of C18 in the original tissue (in µg/g of beef)

Answer 1

Volume of standard solution = 2µL = 2.0x10^-6 L =  2.0x10^-6 Lx (1000 mL / 1L ) = 2.0x10^-3 mL

Mass of C18 in the 2µL of the standard solution = 2.0x10^-3 mL x 0.01 mg/mL = 2.0x10^-5 mg

Hence mass of C18 in the  3 µL of the fame sample = 2.0x10^-5 mg x (14753 / 18334) = 1.61x10^-5 mg

Hence mass of C18 in 50 µL of the FAME sample = 1.61x10^-5 mg x (50 µL / 3 µL) = 2.68x10^-4 mg

50 µL was prepared from 2 mL out of 14 mL of supernatant.

Hence mass of C18 in 5g of beef = 2.68x10^-4 mg x (14 mL / 2 mL ) = 1.876x10^-3 mg = 1.876x10^-6 g

= 1.876 µg C18

Hence concentration of C18 in the original tissue = 1.876 µg C18 / 5 g beef = 0.375 µg / g beef (answer)