Chapter 24, Problem 24.19E

Interpretation Introduction

(a)

Interpretation:

The formulas for the products that are obtained when activated palmitic acid is taken by one turn of the fatty acid spiral are to be stated.

Concept introduction:

A fatty acid is a form of carboxylic acid that contains a long chain of hydrocarbon. It is the principal component of triglyceride and is used for the storage of energy in the body.

The synthesis of fatty acids takes place in the cytoplasm. The starting material of this synthesis is acetyl CoA.

Answer to Problem 24.19E

The formulas for the products that are obtained when activated palmitic acid is taken by one turn of the fatty acid spiral are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}\text{CO}-\text{S}-\text{CoA}+{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+{\text{FADH}}_{\text{2}}+\text{NADH}+{\text{H}}^{+}$

Explanation of Solution

A fatty acid with $16$ carbon atoms is palmitic acid. The thioester product of the reaction that occurs between palmitic acid and HS-CoA is an activated palmitic acid. The reaction for the complete oxidation of activated palmitic acid by the fatty acid reaction spiral is shown below.

$\begin{array}{r}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{14}}\text{CO}-\text{S}-\text{CoA}+7\text{FAD}+7{\text{NAD}}^{+}+7{\text{H}}_{\text{2}}\text{O}+8\text{CoA}-\text{SH}\to \\ {\text{8CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+7{\text{FADH}}_{\text{2}}+7\text{NADH}+7{\text{H}}^{+}\end{array}$

The length of the side chain of the fatty acid is shortens by each trip through the spiral by two carbon atoms. The complete oxidization of palmitic acid having sixteen carbon atoms requires seven trips through the fatty acid oxidation spiral.

When activated palmitic acid is taken through one turn of the fatty acid spiral, one turn shortens the palmitic acid molecule by two carbon atoms. The formation of activated ${\text{C}}_{\text{14}}$ fatty acid occurs with other products. The formulas for the products are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}\text{CO}-\text{S}-\text{CoA}+{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+{\text{FADH}}_{\text{2}}+\text{NADH}+{\text{H}}^{+}$

Conclusion

The formulas for the products through one turn of the fatty acid spiral are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}\text{CO}-\text{S}-\text{CoA}+{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+{\text{FADH}}_{\text{2}}+\text{NADH}+{\text{H}}^{+}$

Interpretation Introduction

(b)

Interpretation:

The formulas for the products that are obtained when activated palmitic acid is taken by two turns of the fatty acid spiral are to be stated.

Concept introduction:

A fatty acid is a form of carboxylic acid that contains a long chain of hydrocarbon. It is the principal component of triglyceride and is used for the storage of energy in the body.

The synthesis of fatty acids takes place in the cytoplasm. The starting material of this synthesis is acetyl CoA.

Answer to Problem 24.19E

The formulas for the products that are obtained when activated palmitic acid is taken by two turns of the fatty acid spiral are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{10}}\text{CO}-\text{S}-\text{CoA}+2{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+2{\text{FADH}}_{\text{2}}+2\text{NADH}+2{\text{H}}^{+}$

Explanation of Solution

When activated palmitic acid is taken through two turns of the fatty acid spiral, one turn shortens the palmitic acid molecule by two carbon atoms. Therefore, two turns shortens the palmitic acid molecule by four carbon atoms. The formation of activated ${\text{C}}_{\text{12}}$ fatty acid occurs with other products. The formulas for the products are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{10}}\text{CO}-\text{S}-\text{CoA}+2{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+2{\text{FADH}}_{\text{2}}+2\text{NADH}+2{\text{H}}^{+}$

Conclusion

The formulas for the products through two turns of the fatty acid spiral are shown below.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{10}}\text{CO}-\text{S}-\text{CoA}+2{\text{CH}}_{\text{3}}\text{CO}-\text{S}-\text{CoA}+2{\text{FADH}}_{\text{2}}+2\text{NADH}+2{\text{H}}^{+}$