4. Palmitic acid is synthesized in vitro with purified substrates and fatty acid

Question

4. Palmitic acid is synthesized in vitro with purified substrates and fatty acid synthase. (If you use standard carbon numbering for fatty acids there is no need to draw the structure of palmitic acid in your answers).

a. If [2-14C]acetyl-CoA is supplied (14CH3-CO-S-CoA), how many carbon(s) in palmitate will be 14C-labeled and at what position(s)? The other substrates in the reaction are not radioactive.

b. If [3-14C]malonyl-CoA (14CO2-CH2-CO-S-CoA) is used in the reaction, how many carbon(s) in palmitate will be 14C-labeled and at what positions? The other substrates in the reaction are not
radioactive.

c. If the malonyl-CoA is supplied with both hydrogens on the malonyl group as deuteriums (CO2-CD2-CO-S-CoA) how many deuterium atoms will be incorporated into palmitate and to which carbons will they be attached? Assume the reaction is performed in a dilute solution of nondeuterated water.

Explanation / Answer

a. As two labeled acetyl molecules are added, two carbons in palmitate will be labeled. They are present in Carbon #s 14 and 16 of palmitate.
The first carbon from the first acetyl molecule is converted to malonyl coA and the second is added to malonyl coA to form acetoacetyl coA.
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b. The labeled C=O group of malonate is released as CO2 in the step of condensation of malonyl coA with acetyl coA. Hence no radiolabeled molecules are seen in palmitate.
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c. If two deuterium ions are present in malonyl coA, during dehydration step , one hydrogen is removed as water. So only one deuterium ion is present in palmitate. It is present in the 14th CH2 group of palmitate.

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