Question

A) Explain how many electrons would enter Complex I of the electron transport chain from the complete oxidation of Myristic Acid, 14:0? Please show work

Answer 1

Complex I is a NADH DEHYDROGENASE which transfer two electron from from NADH and transferred to a lipid-soluble carrier, ubiquinone (UQ).

Note

IN general to remember from complete oxidation

FORMULA: FOR CALCULATING NADH, FADH2, AND ACETYL COA  FOR SATURATED FATTY ACID IN BETA OXIDATION

N= NO. OF CARBON PRESENT IN SATURATED FATTY ACID

FOR NADH AND FADH2 = N/2 -1

FOR ACETYL COA = N/2

IN KREB CYCLE 1 ACETYL COA = 3 NADH +1 FADH

CALCULATION

MYRISTIC ACID = 14 CARBON

NADH = 14/2 - 1 ======= 6

FADH2 = 6

ACETYL COA =14/2=========7

from 7 ACETYL ========7 X 3 = 21 NADH

=========7 X 1 = 7 FADH

NO. OF TOTAL NADH = 6+ 21 =======27

FROM COMPLEX I PASSES 2 ELECTRON FROM 1 NADH SO TOTAL ELECTRON PAS WILL BE 27 X 2 = 54