You analyzed the C18 content of a sample of ground beef. You took 5 g of beef, h

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 mu l. 3 mu l of this were injected onto the GC. The area of the C18 peak was 14753. 2 mu l 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 mu g/g of beef)

Explanation / Answer

concentration of C18 in 2 uL = 0.01 x 0.002 = 2 x 10^-5 mg

let Cx be the concentration of C18 in beef solution,

Cx/(Cx + 2 x 10^-5) = 14753/18334

18334 Cx = 14753 Cx + 0.295

Cx = 0.295/3581 = 8.24 x 10^-5 mg

concentration of C18 in 2 ml = 8.24 x 10^-5 x 0.05/0.003 = 1.37 x 10^-3 mg

concentration in 14 ml original solution = 1.37 x 10^-3 x 14/2 = 9.61 x 10^-3 mg = 9.61 ug

So,

concentration of C18 in original tissue = 9.61 ug/5 g = 1.922 ug/g of beef

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