General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN: 9781285853918
Author: H. Stephen Stoker
Publisher: Cengage Learning
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Chapter 25, Problem 25.106EP

(a)

Interpretation Introduction

Interpretation:

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate succinate has to be determined.

Concept introduction:

Lipogenesis is the process employed for the synthesis of fatty acid. The starting precursor for the synthesis is acetyl CoA. The enzyme employed for the process is fatty acid synthase. It is a multienzyme complex that ties the reaction responsible for the synthesis of fatty acid. This process is the reverse of the degradation of fatty acid.

The Citric acid cycle is a series of biochemical reactions that use acetyl CoA (produced by oxidation of pyruvate) to produce carbon dioxide, NADH and FADH2 in a series of redox reactions.

(a)

Expert Solution
Check Mark

Answer to Problem 25.106EP

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate succinate is butyrate.

Explanation of Solution

Succinic acid is a dicarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the citric acid cycle is a C4 derivative of dicarboxylic acid. The structure of succinic acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  1

Intermediates involved in the lipogenesis are derivatives of C4 molecule butyric acid. Butyric acid is a monocarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the lipogenesis is a C4 derivative of monocarboxylic acid. The structure of butyric acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  2

The structure of succinate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  3

The structure of butyrate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  4

Butyrate and succinate are saturated acid with four carbon atoms in each molecule. butyrate is a monoacid formed as intermediate in lipogenesis while succinate is a diacid formed as intermediate in the citric acid cycle. Therefore, the lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate succinate is butyrate.

(b)

Interpretation Introduction

Interpretation:

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate malate has to be determined.

Concept introduction:

Lipogenesis is the process employed for the synthesis of fatty acid. The starting precursor for the synthesis is acetyl CoA. The enzyme employed for the process is fatty acid synthase. It is a multienzyme complex that ties the reaction responsible for the synthesis of fatty acid. This process is the reverse of the degradation of fatty acid.

The Citric acid cycle is a series of biochemical reactions that use acetyl CoA (produced by oxidation of pyruvate) to produce carbon dioxide, NADH and FADH2 in a series of redox reactions.

Intermediates involved in the lipogenesis are derivative of C4 molecule butyric acid. Butyric acid is a monocarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the lipogenesis is a C4 derivative of monocarboxylic acid. The structure of butyric acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  5

Succinic acid is a dicarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the citric acid cycle is a C4 derivative of dicarboxylic acid. The structure of succinic acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  6

(b)

Expert Solution
Check Mark

Answer to Problem 25.106EP

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate malate is β-hydroxybutyrate.

Explanation of Solution

Malate is the intermediate in the citric acid cycle. The structure of malate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  7

β-Hydroxybutyrate is the intermediate in the lipogenesis. The structure of β-hydroxybutyrate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  8

Malate and β-hydroxybutyrate are the hydroxy derivatives of saturated acid with four carbon atoms in each molecule. Malate is a hydroxy derivative of dicarboxylic acid that is formed as the intermediate in the citric acid cycle whileβ-hydroxybutyrate is a hydroxy derivative of monocarboxylic acid that is formed as the intermediate in the lipogenesis. The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate malate is β-hydroxybutyrate.

(c)

Interpretation Introduction

Interpretation:

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate oxaloacetate has to be determined.

Concept introduction:

Lipogenesis is the process employed for the synthesis of fatty acid. The starting precursor for the synthesis is acetyl CoA. The enzyme employed for the process is fatty acid synthase. It is a multienzyme complex that ties the reaction responsible for the synthesis of fatty acid. This process is the reverse of the degradation of fatty acid.

The Citric acid cycle is a series of biochemical reactions that use acetyl CoA (produced by oxidation of pyruvate) to produce carbon dioxide, NADH and FADH2 in a series of redox reactions.

Intermediates involved in the lipogenesis are derivatives of C4 molecule butyric acid. Butyric acid is a monocarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the lipogenesis is a C4 derivative of monocarboxylic acid. The structure of butyric acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  9

Succinic acid is a dicarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the citric acid cycle is a C4 derivative of dicarboxylic acid. The structure of succinic acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  10

(c)

Expert Solution
Check Mark

Answer to Problem 25.106EP

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate oxaloacetate is acetoacetate.

Explanation of Solution

Oxaloacetate is the intermediate in the citric acid cycle. The structure of oxaloacetate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  11

Acetoacetate is the intermediate in the lipogenesis. The structure of acetoacetate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  12

Oxaloacetate and acetoacetate are the keto derivatives of saturated acid with four carbon atoms in each molecule. Oxaloacetate is a keto derivative of dicarboxylic acid that is formed as the intermediate in the citric acid cycle while acetoacetate is a keto derivative of monocarboxylic acid that is formed as the intermediate in the lipogenesis. Therefore, the lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate oxaloacetate is acetoacetate.

(d)

Interpretation Introduction

Interpretation:

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate fumarate has to be determined.

Concept introduction:

Lipogenesis is the process employed for the synthesis of fatty acid. The starting precursor for the synthesis is acetyl CoA. The enzyme employed for the process is fatty acid synthase. It is a multienzyme complex that ties the reaction responsible for the synthesis of fatty acid. This process is the reverse of the degradation of fatty acid.

The Citric acid cycle is a series of biochemical reactions that use acetyl CoA (produced by oxidation of pyruvate) to produce carbon dioxide, NADH and FADH2 in a series of redox reactions.

Intermediates involved in the lipogenesis are derivatives of C4 molecule butyric acid. Butyric acid is a monocarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the lipogenesis is a C4 derivative of monocarboxylic acid. The structure of butyric acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  13

Succinic acid is a dicarboxylic acid and has 4 carbon atoms. Thus, each intermediate of the citric acid cycle is a C4 derivative of dicarboxylic acid. The structure of succinic acid is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  14

(d)

Expert Solution
Check Mark

Answer to Problem 25.106EP

The lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate fumarate is crotonate.

Explanation of Solution

Fumarate is the intermediate in the citric acid cycle. The structure of fumarate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  15

Crotonate is the intermediate in the lipogenesis. The structure of crotonate is,

General, Organic, and Biological Chemistry, Chapter 25, Problem 25.106EP , additional homework tip  16

Fumarate and crotonate are the unsaturated derivatives of the carboxylic acid with four carbon atoms in each molecule. Fumarate is an unsaturated derivative of dicarboxylic acid that is formed as the citric acid cycle intermediate while crotonate is an unsaturated derivative of monocarboxylic acid that is formed as the lipogenesis intermediate. Therefore, the lipogenesis C4-ACP monoacid intermediate counterpart for the citric acid cycle C4-diacid intermediate fumarate is crotonate.

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Chapter 25 Solutions

General, Organic, and Biological Chemistry

Ch. 25.3 - Prob. 3QQCh. 25.4 - Prob. 1QQCh. 25.4 - Prob. 2QQCh. 25.4 - Prob. 3QQCh. 25.4 - Prob. 4QQCh. 25.4 - Prob. 5QQCh. 25.4 - Prob. 6QQCh. 25.5 - Prob. 1QQCh. 25.5 - Prob. 2QQCh. 25.5 - Prob. 3QQCh. 25.6 - Prob. 1QQCh. 25.6 - Prob. 2QQCh. 25.6 - Prob. 3QQCh. 25.6 - Prob. 4QQCh. 25.6 - Prob. 5QQCh. 25.6 - Prob. 6QQCh. 25.7 - Prob. 1QQCh. 25.7 - Prob. 2QQCh. 25.7 - Prob. 3QQCh. 25.7 - Prob. 4QQCh. 25.7 - The reducing agent needed in the process of...Ch. 25.7 - Prob. 6QQCh. 25.8 - Prob. 1QQCh. 25.8 - Prob. 2QQCh. 25.9 - Prob. 1QQCh. 25.9 - Prob. 2QQCh. 25.9 - Prob. 3QQCh. 25.9 - Prob. 4QQCh. 25.10 - Which of the following substances cannot be...Ch. 25.10 - Prob. 2QQCh. 25.10 - Which of the following processes occurs within the...Ch. 25.11 - Prob. 1QQCh. 25.11 - Prob. 2QQCh. 25.11 - Prob. 3QQCh. 25 - Indicate whether each of the following aspects of...Ch. 25 - Indicate whether each of the following aspects of...Ch. 25 - Indicate whether each of the following pairings of...Ch. 25 - Prob. 25.4EPCh. 25 - Indicate whether each of the following statements...Ch. 25 - Prob. 25.6EPCh. 25 - Prob. 25.7EPCh. 25 - What is a chylomicron?Ch. 25 - What are the products of the complete hydrolysis...Ch. 25 - What are the major products of the incomplete...Ch. 25 - Prob. 25.11EPCh. 25 - At what location are free fatty acids and...Ch. 25 - Prob. 25.13EPCh. 25 - Prob. 25.14EPCh. 25 - Prob. 25.15EPCh. 25 - Prob. 25.16EPCh. 25 - Prob. 25.17EPCh. 25 - Prob. 25.18EPCh. 25 - Prob. 25.19EPCh. 25 - Prob. 25.20EPCh. 25 - Prob. 25.21EPCh. 25 - Prob. 25.22EPCh. 25 - Prob. 25.23EPCh. 25 - Prob. 25.24EPCh. 25 - Prob. 25.25EPCh. 25 - Prob. 25.26EPCh. 25 - Prob. 25.27EPCh. 25 - Identify the oxidizing agent needed in Step 3 of a...Ch. 25 - Prob. 25.29EPCh. 25 - Prob. 25.30EPCh. 25 - Prob. 25.31EPCh. 25 - Prob. 25.32EPCh. 25 - Prob. 25.33EPCh. 25 - Prob. 25.34EPCh. 25 - Prob. 25.35EPCh. 25 - Prob. 25.36EPCh. 25 - Prob. 25.37EPCh. 25 - Prob. 25.38EPCh. 25 - Prob. 25.39EPCh. 25 - Prob. 25.40EPCh. 25 - Prob. 25.41EPCh. 25 - Prob. 25.42EPCh. 25 - How many turns of the -oxidation pathway would be...Ch. 25 - How many turns of the -oxidation pathway would be...Ch. 25 - Prob. 25.45EPCh. 25 - Prob. 25.46EPCh. 25 - Prob. 25.47EPCh. 25 - Prob. 25.48EPCh. 25 - Prob. 25.49EPCh. 25 - Explain why fatty acids cannot serve as fuel for...Ch. 25 - Prob. 25.51EPCh. 25 - Prob. 25.52EPCh. 25 - Prob. 25.53EPCh. 25 - Prob. 25.54EPCh. 25 - Prob. 25.55EPCh. 25 - Prob. 25.56EPCh. 25 - Prob. 25.57EPCh. 25 - Prob. 25.58EPCh. 25 - Prob. 25.59EPCh. 25 - Prob. 25.60EPCh. 25 - Prob. 25.61EPCh. 25 - Why does a deficiency of carbohydrates in the diet...Ch. 25 - Prob. 25.63EPCh. 25 - Prob. 25.64EPCh. 25 - Prob. 25.65EPCh. 25 - Prob. 25.66EPCh. 25 - Prob. 25.67EPCh. 25 - Prob. 25.68EPCh. 25 - Prob. 25.69EPCh. 25 - Prob. 25.70EPCh. 25 - Prob. 25.71EPCh. 25 - Prob. 25.72EPCh. 25 - Prob. 25.73EPCh. 25 - Prob. 25.74EPCh. 25 - Prob. 25.75EPCh. 25 - Severe ketosis situations produce acidosis....Ch. 25 - Prob. 25.77EPCh. 25 - Prob. 25.78EPCh. 25 - Prob. 25.79EPCh. 25 - Prob. 25.80EPCh. 25 - Prob. 25.81EPCh. 25 - Prob. 25.82EPCh. 25 - Prob. 25.83EPCh. 25 - Prob. 25.84EPCh. 25 - Prob. 25.85EPCh. 25 - Prob. 25.86EPCh. 25 - Prob. 25.87EPCh. 25 - Prob. 25.88EPCh. 25 - Prob. 25.89EPCh. 25 - Prob. 25.90EPCh. 25 - Prob. 25.91EPCh. 25 - Prob. 25.92EPCh. 25 - Prob. 25.93EPCh. 25 - Prob. 25.94EPCh. 25 - What role does molecular oxygen, O2, play in fatty...Ch. 25 - Prob. 25.96EPCh. 25 - Prob. 25.97EPCh. 25 - Prob. 25.98EPCh. 25 - Prob. 25.99EPCh. 25 - Prob. 25.100EPCh. 25 - Prob. 25.101EPCh. 25 - Prob. 25.102EPCh. 25 - Prob. 25.103EPCh. 25 - Prob. 25.104EPCh. 25 - Prob. 25.105EPCh. 25 - Prob. 25.106EPCh. 25 - Prob. 25.107EPCh. 25 - Prob. 25.108EPCh. 25 - Prob. 25.109EPCh. 25 - Prob. 25.110EPCh. 25 - Prob. 25.111EPCh. 25 - Prob. 25.112EPCh. 25 - Prob. 25.113EPCh. 25 - Prob. 25.114EP
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