Chapter 23, Problem 8P

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the oxidation to CO₂ and water of myristic acid should be written.

Concept Introduction:

The pathway of ß oxidation of saturated fatty acids involves repetitive four steps. The first three steps are to create a carbonyl group on ß carbon by oxidizing the bond between a and ß carbon. The resulted olefin is subsequently subjected to hydration and oxidation. In the fourth step ß keto ester is cleaved in a reverse Claisen condensation reaction, leaving an acetate unit and a fatty acid chain that lacks two carbons than it had. This shorter fatty acid chain can again participate in another ß oxidation cycle. The acetyl CoA produced is further metabolized in TCA cycle and amino acid biosynthesis.

Explanation of Solution

Myristic acid is a saturated fatty acid. First myristic acid is converted into a CoA derivative before the ß oxidation.

  ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}\text{COOH + ATP + CoASH }\to {\text{ CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}{\text{CO-S-CoA + AMP + PP}}_{\text{i}}$

Then after six circles of ß oxidation myristoyl CoA is converted into 7 acetyl CoA units.

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{12}}{\text{CO-S-CoA + 6 CoA-SH + 6 H}}_2{\text{O + 6 NAD}}^{+}\text{ + 6 FAD }\to \\ {\text{ 7 CH}}_{\text{3}}{\text{CO-S-CoA + 6 NADH + 6 H}}^{+}{\text{ + 6 FADH}}_2\end{array}$

Then the acetyl CoA produced enter into TCA cycle

  $\begin{array}{l}{\text{7 CH}}_{\text{3}}{\text{CO-S-CoA + 21 H}}_2{\text{O + 21 NAD}}^{+}\text{ + 7 FAD }\to \\ {\text{ 14 CO}}_2{\text{ + 7 CoA-SH + 21 NADH + 21 H}}^{+}{\text{ + 7 FADH}}_2\end{array}$

Above reaction is achieved by GDP phosphorylation. It is equivalent to

  $7{\text{ ADP + 7 P}}_i\to {\text{ 7 ATP + 7 H}}_2\text{O}$

NAD+ is recycled in electron transport chain

  ${\text{27 NADH + 27 H}}^{+}\text{ + 13}{\text{.5 O}}_2\to {\text{ 27 NAD}}^{+}{\text{ + 27 H}}_2\text{O}$

Above reaction supports the production of $2.5\times 27({\text{ ADP + P}}_i\to {\text{ ATP + H}}_2\text{O ) }$

FAD is recycled by $13{\text{ FADH}}_2\text{ + 6}{\text{.5 O}}_2\to {\text{ 13 FAD + 13 H}}_2\text{O}$

Above reaction supports the production of $\text{1}\text{.5 }\times \text{ 13 }\left({\text{ADP + P}}_i\to {\text{ ATP + H}}_2\text{O}\right)$

AMP produced is phosphorylated to ADP using ATP and PP_i is hydrolyzed.

  ${\text{AMP + ATP + PP}}_i{\text{ + H}}_2\text{O }\to {\text{ 2 ADP + 2 P}}_i$

Therefore the overall equation will be

  ${\text{CH}}_3{\left({\text{CH}}_2\right)}_{12}{\text{COOH + 92 ADP + 92 P}}_i{\text{ + 20 O}}_2\to {\text{ 92 ATP + 14 CO}}_2{\text{ + 106 H}}_2\text{O}$

Interpretation Introduction

(b)

Interpretation The balanced equation for the oxidation to CO₂ and water of stearic acid should be written.

Concept Introduction:

The pathway of ß oxidation of saturated fatty acids involves repetitive four steps. The first three steps are to create a carbonyl group on ß carbon by oxidizing the bond between a and ß carbon. The resulted olefin is subsequently subjected to hydration and oxidation. In the fourth step ß keto ester is cleaved in a reverse Claisen condensation reaction, leaving an acetate unit and a fatty acid chain that lacks two carbons than it had. This shorter fatty acid chain can again participate in another ß oxidation cycle. The acetyl CoA produced is further metabolized in TCA cycle and amino acid biosynthesis.

Explanation of Solution

Stearic acid is a saturated fatty acid. First stearic acid is converted into a CoA derivative before the ß oxidation.

  ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{16}}\text{COOH + ATP + CoASH }\to {\text{ CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{16}}{\text{CO-S-CoA + AMP + PP}}_{\text{i}}$

Then after eight circles of ß oxidation stearoyl CoA is converted into 9 acetyl CoA units.

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{16}}{\text{CO-S-CoA + 8 CoA-SH + 8 H}}_2{\text{O + 8 NAD}}^{+}\text{ + 8 FAD }\to \\ {\text{ 9 CH}}_{\text{3}}{\text{CO-S-CoA + 8 NADH + 8 H}}^{+}{\text{ + 8 FADH}}_2\end{array}$

Then the acetyl CoA produced enter into TCA cycle

  $\begin{array}{l}{\text{9 CH}}_{\text{3}}{\text{CO-S-CoA + 27 H}}_2{\text{O + 27 NAD}}^{+}\text{ + 9 FAD }\to \\ {\text{ 18 CO}}_2{\text{ + 9 CoA-SH + 27 NADH + 27 H}}^{+}{\text{ + 9 FADH}}_2\end{array}$

Above reaction is achieved by GDP phosphorylation. It is equivalent to

  ${\text{9 ADP + 9 P}}_i\to {\text{ 9 ATP + 9 H}}_2\text{O}$

NAD+ is recycled in electron transport chain

  ${\text{35 NADH + 35 H}}^{+}\text{ + 17}{\text{.5 O}}_2\to {\text{ 35 NAD}}^{+}{\text{ + 35 H}}_2\text{O}$

Above reaction supports the production of $2.5\times 35({\text{ ADP + P}}_i\to {\text{ ATP + H}}_2\text{O ) }$

FAD is recycled by $17{\text{ FADH}}_2\text{ + 8}{\text{.5 O}}_2\to {\text{ 17 FAD + 17 H}}_2\text{O}$

Above reaction supports the production of $\text{1}\text{.5 }\times \text{ 17 }\left({\text{ADP + P}}_i\to {\text{ ATP + H}}_2\text{O}\right)$

AMP produced is phosphorylated to ADP using ATP and PP_i is hydrolyzed.

  ${\text{AMP + ATP + PP}}_i{\text{ + H}}_2\text{O }\to {\text{ 2 ADP + 2 P}}_i$

Therefore the overall equation will be

  ${\text{CH}}_3{\left({\text{CH}}_2\right)}_{16}{\text{COOH + 120 ADP + 120 P}}_i{\text{ + 26 O}}_2\to {\text{ 120 ATP + 18 CO}}_2{\text{ + 138 H}}_2\text{O}$

Interpretation Introduction

(c)

Interpretation:

The balanced equation for the oxidation to CO₂ and water of a-linolenic acid should be written.

Concept Introduction:

Unsaturated fatty acids are also subjected to ß oxidation. But for this two additional enzymes; an isomerase and a reductase are essential to manipulate cis double bonds of fatty acid. As the first steps for monounsaturated fatty acids like oleic acid which is a 18 carbon fatty acid with one double bond at 9,10 position., it normally undergoes the ß oxidation leaving 3 acetyl CoA and the cis-cis-Δ³-dodecenoyl-CoA product. This intermediate is not a substrate for acyl CoA dehydrogenase. This intermediate is then subjected to enoyl-CoA isomerase enzyme activity which rearranges the cis-Δ³ double bond to a trans-Δ² double bond. This intermediate with trans-Δ² double bond is preceded via normal ß oxidation.

But for poly unsaturated fatty acids like linoleic acid, ß oxidation occurs through three cycles and the enoyl CoA product is subjected to enoyl-CoA isomerase permitting another round of ß oxidation. The resulting cis-Δ⁴ enoyl CoA is converted normally to trans-Δ², cis-Δ⁴ species by acyl CoA dehydrogenase. This product is not a substrate for enoyl CoA hydratase. In mammals 2,4-dienoyl CoA reductase produces a trans-Δ³ enoyl product which is then converted to the trans-Δ² CoA by an enoyl CoA isomerase. This product then can normally participate in the ß oxidation.

Explanation of Solution

a linolenic acid is a polyunsaturated C-18 fatty acid with three double bonds at C-9, C-12 and C-15 positions. Eight rounds of ß oxidation will produce 9 molecules of acetyl CoA, without the reduction of FAD into FADH₂ in two steps. Therefore two fewer molecules of FADH₂ will enter into electron transport chain than in the case of stearic acid and 1 fewer molecule of O₂ will be consumed. Therefore amount of ATP is reduced by 3, because $1.5\times 2=3$ . Furthermore NADH will be consumed in the fifth cycle of ß oxidation to resolve conjugated double bond. This makes amount of ATP produced reduced by 2.5 and O₂ consumed reduced by 0.5 than in case of stearic acid.

Therefore the overall equation will be

  ${\text{C}}_{17}{\text{H}}_{29}\text{COOH + 113}\text{.5 ADP + 113}{\text{.5 P}}_i\text{ + 24}{\text{.5 O}}_2\to \text{ 113}{\text{.5 ATP + 18 CO}}_2\text{ + 128}{\text{.5 H}}_2\text{O}$

Interpretation Introduction

(d)

Interpretation:

The balanced equation for the oxidation to CO₂ and water of a-linolenic acid should be written.

Concept Introduction:

Unsaturated fatty acids are also subjected to ß oxidation. But for this two additional enzymes; an isomerase and a reductase are essential to manipulate cis double bonds of fatty acid. As the first steps for monounsaturated fatty acids like oleic acid which is a 18 carbon fatty acid with one double bond at 9,10 position., it normally undergoes the ß oxidation leaving 3 acetyl CoA and the cis-cis-Δ³-dodecenoyl-CoA product. This intermediate is not a substrate for acyl CoA dehydrogenase. This intermediate is then subjected to enoyl-CoA isomerase enzyme activity which rearranges the cis-Δ³ double bond to a trans-Δ² double bond. This intermediate with trans-Δ² double bond is preceded via normal ß oxidation.

But for poly unsaturated fatty acids like linoleic acid, ß oxidation occurs through three cycles and the enoyl CoA product is subjected to enoyl-CoA isomerase permitting another round of ß oxidation. The resulting cis-Δ⁴ enoyl CoA is converted normally to trans-Δ², cis-Δ⁴ species by acyl CoA dehydrogenase. This product is not a substrate for enoyl CoA hydratase. In mammals 2,4-dienoyl CoA reductase produces a trans-Δ³ enoyl product which is then converted to the trans-Δ² CoA by an enoyl CoA isomerase. This product then can normally participate in the ß oxidation.

Explanation of Solution

Arachidonic acid is a polyunsaturated C-20 fatty acid with four double bonds at C-5, C-8, C-11 and C-14 positions. 9 rounds of ß oxidation will produce 10 molecules of acetyl CoA, without the reduction of FAD into FADH₂ in two steps. Therefore two fewer molecules of FADH₂ will enter into electron transport chain than in the case of stearic acid. Furthermore two NADH molecules will be consumed to resolve two conjugated double bonds.

Therefore the overall equation will be

  ${\text{C}}_{19}{\text{H}}_{31}{\text{COOH + 125 ADP + 125 P}}_i{\text{ + 27 O}}_2\to {\text{ 125 ATP + 20 CO}}_2{\text{ + 141 H}}_2\text{O}$