GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP
GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP
7th Edition
ISBN: 9781305866966
Author: STOKER
Publisher: CENGAGE L
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Chapter 25, Problem 25.105EP

(a)

Interpretation Introduction

Interpretation:

The citric acid cycle diacid intermediate counterpart for lipogenesis C4-ACP monoacid intermediate butyrate 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 25.105EP

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

Explanation of Solution

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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  1

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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  2

The structure of butyrate is,

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  3

The structure of succinate is,

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  4

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

(b)

Interpretation Introduction

Interpretation:

The citric acid cycle diacid intermediate counterpart for lipogenesis C4-ACP monoacid intermediate acetoacetate 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  6

(b)

Expert Solution
Check Mark

Answer to Problem 25.105EP

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

Explanation of Solution

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  7

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  8

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

(c)

Interpretation Introduction

Interpretation:

The citric acid cycle diacid intermediate counterpart for lipogenesis C4-ACP monoacid intermediate β-hydroxybutyrate 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  10

(c)

Expert Solution
Check Mark

Answer to Problem 25.105EP

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

Explanation of Solution

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  11

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  12

β-Hydroxybutyrate and malate are the hydroxy derivatives of saturated acid with four carbon atoms in each molecule. β-Hydroxybutyrate is a hydroxy derivative of monocarboxylic acid while malate is a hydroxy derivative of dicarboxylic acid. Therefore, the citric acid cycle diacid intermediate counterpart for lipogenesis C4-ACP monoacid β-hydroxybutyrate intermediate is malate.

(d)

Interpretation Introduction

Interpretation:

The citric acid cycle diacid intermediate counterpart for lipogenesis C4-ACP monoacid intermediate crotonate 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , 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,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  14

(d)

Expert Solution
Check Mark

Answer to Problem 25.105EP

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

Explanation of Solution

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  15

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

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP, Chapter 25, Problem 25.105EP , additional homework tip  16

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

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Chapter 25 Solutions

GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP

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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